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For part 2:

With \begin{equation*} y_{\alpha_1}=\begin{pmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix},y_{\alpha_2}=\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{pmatrix},y_{\alpha_3}=\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix} \end{equation*} we have \begin{align} [y_{\alpha_1},y_{\alpha_2}]&= 0, & (1)\\ [y_{\alpha_1},y_{\alpha_3}]&= -y_{\alpha_2}. & (2) \end{align}

Then \begin{align} 0 &=\delta_{21}\circ \delta_3(v_{(−2,0,2)})=\delta_{21}(a_1y_{\alpha_3}v_{(−2,1,1)},a_2y_{\alpha_1}v_{(−1,−1,−2)})\\ &=(a_1y_{\alpha_3}b_1y_{\alpha_1}^2+a_2y_{\alpha_1}b_2(2y_{\alpha_1}y_{\alpha_3}+y_{\alpha_2}))v_{(0,-1,1)} \\ &=(a_1b_1y_{\alpha_3}y_{\alpha_1}^2+ 2a_2b_2y_{\alpha_1}^2y_{\alpha_3}+a_2b_2y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)}\\ \end{align} with $(2)$ follows \begin{align} &=(a_1b_1(y_{\alpha_1}y_{\alpha_3}+y_{\alpha_2})y_{\alpha_1}+2a_2b_2y_{\alpha_1}^2y_{\alpha_3}+a_2b_2y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)} \\ &=(a_1b_1y_{\alpha_1}y_{\alpha_3}y_{\alpha_1}+a_1b_1y_{\alpha_2}y_{\alpha_1}+2a_2b_2y_{\alpha_1}^2y_{\alpha_3}+a_2b_2y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)} \end{align} Applying $(2)$ again and additionally $(1)$ we get \begin{align*} &=(a_1b_1y_{\alpha_1}(y_{\alpha_1}y_{\alpha_3}+y_{\alpha_2})+a_1b_1y_{\alpha_1}y_{\alpha_2}+2a_2b_2y_{\alpha_1}^2y_{\alpha_3}+a_2b_2y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)} \\ &=(a_1b_1y_{\alpha_1}^2y_{\alpha_3}+a_1b_1y_{\alpha_1}y_{\alpha_2}+a_1b_1y_{\alpha_1}y_{\alpha_2}+2a_2b_2y_{\alpha_1}^2y_{\alpha_3}+a_2b_2y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)} \\ &=(a_1b_1y_{\alpha_1}^2y_{\alpha_3}+2a_1b_1y_{\alpha_1}y_{\alpha_2}+2a_2b_2y_{\alpha_1}^2y_{\alpha_3}+a_2b_2y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)} \\ &=((a_1b_1+2a_2b_2)y_{\alpha_1}^2y_{\alpha_3}+(2a_1b_1+a_2b_2)y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)} . \end{align*}

But this would again imply that $a_i, b_i$ are trivial?! So which mistake I did this time...

For part 2:

With \begin{equation*} y_{\alpha_1}=\begin{pmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix},y_{\alpha_2}=\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{pmatrix},y_{\alpha_3}=\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix} \end{equation*} we have \begin{align} [y_{\alpha_1},y_{\alpha_2}]&= 0, & (1)\\ [y_{\alpha_1},y_{\alpha_3}]&= -y_{\alpha_2}. & (2) \end{align}

Then \begin{align} 0 &=\delta_{21}\circ \delta_3(v_{(−2,0,2)})=\delta_{21}(a_1y_{\alpha_3}v_{(−2,1,1)},a_2y_{\alpha_1}v_{(−1,−1,−2)})\\ &=(a_1y_{\alpha_3}b_1y_{\alpha_1}^2+a_2y_{\alpha_1}b_2(y_{\alpha_1}y_{\alpha_3}+2y_{\alpha_2}))v_{(0,-1,1)} \\ &=(a_1b_1y_{\alpha_3}y_{\alpha_1}^2+ a_2b_2y_{\alpha_1}^2y_{\alpha_3}+2a_2b_2y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)}\\ \end{align} with $(2)$ follows \begin{align} &=(a_1b_1(y_{\alpha_1}y_{\alpha_3}+y_{\alpha_2})y_{\alpha_1}+a_2b_2y_{\alpha_1}^2y_{\alpha_3}+2a_2b_2y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)} \\ &=(a_1b_1y_{\alpha_1}y_{\alpha_3}y_{\alpha_1}+a_1b_1y_{\alpha_2}y_{\alpha_1}+a_2b_2y_{\alpha_1}^2y_{\alpha_3}+2a_2b_2y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)} \end{align} Applying $(2)$ again and additionally $(1)$ we get \begin{align*} &=(a_1b_1y_{\alpha_1}(y_{\alpha_1}y_{\alpha_3}+y_{\alpha_2})+a_1b_1y_{\alpha_1}y_{\alpha_2}+a_2b_2y_{\alpha_1}^2y_{\alpha_3}+2a_2b_2y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)} \\ &=(a_1b_1y_{\alpha_1}^2y_{\alpha_3}+a_1b_1y_{\alpha_1}y_{\alpha_2}+a_1b_1y_{\alpha_1}y_{\alpha_2}+a_2b_2y_{\alpha_1}^2y_{\alpha_3}+2a_2b_2y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)} \\ &=(a_1b_1y_{\alpha_1}^2y_{\alpha_3}+2a_1b_1y_{\alpha_1}y_{\alpha_2}+a_2b_2y_{\alpha_1}^2y_{\alpha_3}+2a_2b_2y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)} \\ &=((a_1b_1+a_2b_2)y_{\alpha_1}^2y_{\alpha_3}+(2a_1b_1+2a_2b_2)y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)}. \end{align*}

and $a_1b_1=-a_2b_2$ is enough. As in [H| mentioned it is possible to choose $a_i,b_i \in \{-1,1\}$.

For part 2:

With \begin{equation*} y_{\alpha_1}=\begin{pmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix},y_{\alpha_2}=\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{pmatrix},y_{\alpha_3}=\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix} \end{equation*} we have \begin{align} [y_{\alpha_1},y_{\alpha_2}]&= 0, & (1)\\ [y_{\alpha_1},y_{\alpha_3}]&= -y_{\alpha_2}. & (2) \end{align}

Then \begin{align} 0 &=\delta_{21}\circ \delta_3(v_{(−2,0,2)})=\delta_{21}(a_1y_{\alpha_3}v_{(−2,1,1)},a_2y_{\alpha_1}v_{(−1,−1,−2)})\\ &=(a_1y_{\alpha_3}b_1y_{\alpha_1}^2+a_2y_{\alpha_1}b_2(2y_{\alpha_1}y_{\alpha_3}+y_{\alpha_2}))v_{(0,-1,1)} \\ &=(a_1b_1y_{\alpha_3}y_{\alpha_1}^2+ 2a_2b_2y_{\alpha_1}^2y_{\alpha_3}+a_2b_2y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)}\\ \end{align} with $(1)$ follows \begin{align} &=(a_1b_1(y_{\alpha_1}y_{\alpha_3}+y_{\alpha_2})y_{\alpha_1}+2a_2b_2y_{\alpha_1}^2y_{\alpha_3}+a_2b_2y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)} \\ &=(a_1b_1y_{\alpha_1}y_{\alpha_3}y_{\alpha_1}+a_1b_1y_{\alpha_2}y_{\alpha_1}+2a_2b_2y_{\alpha_1}^2y_{\alpha_3}+a_2b_2y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)} \end{align} Applying $(1)$ again and additionally $(2)$ we get \begin{align*} &=(a_1b_1y_{\alpha_1}(y_{\alpha_1}y_{\alpha_3}+y_{\alpha_2})+a_1b_1y_{\alpha_1}y_{\alpha_2}+2a_2b_2y_{\alpha_1}^2y_{\alpha_3}+a_2b_2y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)} \\ &=(a_1b_1y_{\alpha_1}^2y_{\alpha_3}+a_1b_1y_{\alpha_1}y_{\alpha_2}+a_1b_1y_{\alpha_1}y_{\alpha_2}+2a_2b_2y_{\alpha_1}^2y_{\alpha_3}+a_2b_2y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)} \\ &=(a_1b_1y_{\alpha_1}^2y_{\alpha_3}+2a_1b_1y_{\alpha_1}y_{\alpha_2}+2a_2b_2y_{\alpha_1}^2y_{\alpha_3}+a_2b_2y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)} \\ &=((a_1b_1+2a_2b_2)y_{\alpha_1}^2y_{\alpha_3}+(2a_1b_1+a_2b_2)y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)} . \end{align*}

But this would again imply that $a_i, b_i$ are trivial?! So which mistake I did this time...

For part 2:

With \begin{equation*} y_{\alpha_1}=\begin{pmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix},y_{\alpha_2}=\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{pmatrix},y_{\alpha_3}=\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix} \end{equation*} we have \begin{align} [y_{\alpha_1},y_{\alpha_2}]&= 0, & (1)\\ [y_{\alpha_1},y_{\alpha_3}]&= -y_{\alpha_2}. & (2) \end{align}

Then \begin{align} 0 &=\delta_{21}\circ \delta_3(v_{(−2,0,2)})=\delta_{21}(a_1y_{\alpha_3}v_{(−2,1,1)},a_2y_{\alpha_1}v_{(−1,−1,−2)})\\ &=(a_1y_{\alpha_3}b_1y_{\alpha_1}^2+a_2y_{\alpha_1}b_2(2y_{\alpha_1}y_{\alpha_3}+y_{\alpha_2}))v_{(0,-1,1)} \\ &=(a_1b_1y_{\alpha_3}y_{\alpha_1}^2+ 2a_2b_2y_{\alpha_1}^2y_{\alpha_3}+a_2b_2y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)}\\ \end{align} with $(2)$ follows \begin{align} &=(a_1b_1(y_{\alpha_1}y_{\alpha_3}+y_{\alpha_2})y_{\alpha_1}+2a_2b_2y_{\alpha_1}^2y_{\alpha_3}+a_2b_2y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)} \\ &=(a_1b_1y_{\alpha_1}y_{\alpha_3}y_{\alpha_1}+a_1b_1y_{\alpha_2}y_{\alpha_1}+2a_2b_2y_{\alpha_1}^2y_{\alpha_3}+a_2b_2y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)} \end{align} Applying $(2)$ again and additionally $(1)$ we get \begin{align*} &=(a_1b_1y_{\alpha_1}(y_{\alpha_1}y_{\alpha_3}+y_{\alpha_2})+a_1b_1y_{\alpha_1}y_{\alpha_2}+2a_2b_2y_{\alpha_1}^2y_{\alpha_3}+a_2b_2y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)} \\ &=(a_1b_1y_{\alpha_1}^2y_{\alpha_3}+a_1b_1y_{\alpha_1}y_{\alpha_2}+a_1b_1y_{\alpha_1}y_{\alpha_2}+2a_2b_2y_{\alpha_1}^2y_{\alpha_3}+a_2b_2y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)} \\ &=(a_1b_1y_{\alpha_1}^2y_{\alpha_3}+2a_1b_1y_{\alpha_1}y_{\alpha_2}+2a_2b_2y_{\alpha_1}^2y_{\alpha_3}+a_2b_2y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)} \\ &=((a_1b_1+2a_2b_2)y_{\alpha_1}^2y_{\alpha_3}+(2a_1b_1+a_2b_2)y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)} . \end{align*}

But this would again imply that $a_i, b_i$ are trivial?! So which mistake I did this time...

For part 2::

With \begin{equation*} y_{\alpha_1}=\begin{pmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix},y_{\alpha_2}=\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{pmatrix},y_{\alpha_3}=\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix} \end{equation*} we have \begin{align} [y_{\alpha_1},y_{\alpha_2}]&= 0, & (1)\\ [y_{\alpha_1},y_{\alpha_3}]&= -y_{\alpha_2}. & (2) \end{align}

Then \begin{align} 0 &=\delta_{21}\circ \delta_3(v_{(−2,0,2)})=\delta_{21}(a_1y_{\alpha_3}v_{(−2,1,1)},a_2y_{\alpha_1}v_{(−1,−1,−2)})\\ &=(a_1y_{\alpha_3}b_1y_{\alpha_1}^2+a_2y_{\alpha_1}b_2(2y_{\alpha_1}y_{\alpha_3}+y_{\alpha_2}))v_{(0,-1,1)} \\ &=(a_1b_1y_{\alpha_3}y_{\alpha_1}^2+ 2a_2b_2y_{\alpha_1}^2y_{\alpha_3}+a_2b_2y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)}\\ \end{align} with $(1)$ follows \begin{align} &=(a_1b_1(y_{\alpha_1}y_{\alpha_3}+y_{\alpha_2})y_{\alpha_1}+2a_2b_2y_{\alpha_1}^2y_{\alpha_3}+a_2b_2y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)} \\ &=(a_1b_1y_{\alpha_1}y_{\alpha_3}y_{\alpha_1}+a_1b_1y_{\alpha_2}y_{\alpha_1}+2a_2b_2y_{\alpha_1}^2y_{\alpha_3}+a_2b_2y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)} \end{align} Applying $(1)$ again and additionally $(2)$ we get \begin{align*} &=(a_1b_1y_{\alpha_1}(y_{\alpha_1}y_{\alpha_3}+y_{\alpha_2})+a_1b_1y_{\alpha_1}y_{\alpha_2}+2a_2b_2y_{\alpha_1}^2y_{\alpha_3}+a_2b_2y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)} \\ &=(a_1b_1y_{\alpha_1}^2y_{\alpha_3}+a_1b_1y_{\alpha_1}y_{\alpha_2}+a_1b_1y_{\alpha_1}y_{\alpha_2}+2a_2b_2y_{\alpha_1}^2y_{\alpha_3}+a_2b_2y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)} \\ &=(a_1b_1y_{\alpha_1}^2y_{\alpha_3}+2a_1b_1y_{\alpha_1}y_{\alpha_2}+2a_2b_2y_{\alpha_1}^2y_{\alpha_3}+a_2b_2y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)} \\ &=((a_1b_1+2a_2b_2)y_{\alpha_1}^2y_{\alpha_3}+(2a_1b_1+a_2b_2)y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)} . \end{align*}

But this would again imply that $a_i, b_i$ are trivial?!

For part 2:

With \begin{equation*} y_{\alpha_1}=\begin{pmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix},y_{\alpha_2}=\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{pmatrix},y_{\alpha_3}=\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix} \end{equation*} we have \begin{align} [y_{\alpha_1},y_{\alpha_2}]&= 0, & (1)\\ [y_{\alpha_1},y_{\alpha_3}]&= -y_{\alpha_2}. & (2) \end{align}

Then \begin{align} 0 &=\delta_{21}\circ \delta_3(v_{(−2,0,2)})=\delta_{21}(a_1y_{\alpha_3}v_{(−2,1,1)},a_2y_{\alpha_1}v_{(−1,−1,−2)})\\ &=(a_1y_{\alpha_3}b_1y_{\alpha_1}^2+a_2y_{\alpha_1}b_2(2y_{\alpha_1}y_{\alpha_3}+y_{\alpha_2}))v_{(0,-1,1)} \\ &=(a_1b_1y_{\alpha_3}y_{\alpha_1}^2+ 2a_2b_2y_{\alpha_1}^2y_{\alpha_3}+a_2b_2y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)}\\ \end{align} with $(1)$ follows \begin{align} &=(a_1b_1(y_{\alpha_1}y_{\alpha_3}+y_{\alpha_2})y_{\alpha_1}+2a_2b_2y_{\alpha_1}^2y_{\alpha_3}+a_2b_2y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)} \\ &=(a_1b_1y_{\alpha_1}y_{\alpha_3}y_{\alpha_1}+a_1b_1y_{\alpha_2}y_{\alpha_1}+2a_2b_2y_{\alpha_1}^2y_{\alpha_3}+a_2b_2y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)} \end{align} Applying $(1)$ again and additionally $(2)$ we get \begin{align*} &=(a_1b_1y_{\alpha_1}(y_{\alpha_1}y_{\alpha_3}+y_{\alpha_2})+a_1b_1y_{\alpha_1}y_{\alpha_2}+2a_2b_2y_{\alpha_1}^2y_{\alpha_3}+a_2b_2y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)} \\ &=(a_1b_1y_{\alpha_1}^2y_{\alpha_3}+a_1b_1y_{\alpha_1}y_{\alpha_2}+a_1b_1y_{\alpha_1}y_{\alpha_2}+2a_2b_2y_{\alpha_1}^2y_{\alpha_3}+a_2b_2y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)} \\ &=(a_1b_1y_{\alpha_1}^2y_{\alpha_3}+2a_1b_1y_{\alpha_1}y_{\alpha_2}+2a_2b_2y_{\alpha_1}^2y_{\alpha_3}+a_2b_2y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)} \\ &=((a_1b_1+2a_2b_2)y_{\alpha_1}^2y_{\alpha_3}+(2a_1b_1+a_2b_2)y_{\alpha_1}y_{\alpha_2})v_{(0,-1,1)} . \end{align*}

But this would again imply that $a_i, b_i$ are trivial?! So which mistake I did this time...