The answer for this question can be concluded from my answer on this question:
About some property of automorphism of octonions

Does it use Lie theory ? We have used the fact that orthogonal mapping on $\mathbb R^7$ has fixed point. This fact can be proved using linear algebra, i believe.

It remains to prove that sum of the angles is $\pi$. I thought that sum of the angles should be zero.

To prove this we need to consider basis $\langle 1, i, u, iu \rangle$ and products of these by $v$ (notation from my answer to the other question; $i,u,v$ are perpendicular imaginary unit octonions and $v$ is perpendicular also to $iu$).

**EDIT 2017-08-20**

One idea is to use following formula for octonion multiplication. Let's define octonions as pairs $(a,\mathbf v)$ where $a$ is complex number and $\mathbf v$ vector in $\Bbb C^3$. Then octonion multiplication can be defined as $$(a,\mathbf v)(b,\mathbf w)=(ab-\mathbf {v\cdot w},a\mathbf w+\bar b \mathbf v + \mathbf {v \times w})$$
It can be proved by applying twice Cayley-Dickson formula and complex conjugation to last coefficient. BTW, I have placed this formula already in one question on MO, but it was closed by administrators. I would like to have this formula on MO, so I don't forget it.

We can see that complex cross product is preserved by $SU_3$.

Second idea is to observe that subgroup of $G_2$ which fix $i$ is generated by set $M=\{(L_iR_x)^2:x \text{ imaginary perpendicular to i}\}$.

By $L_i$ I denote matrix of left multiplication by octonion $i$, $R_x$ denote right multiplication by octonion $x$. Element $(L_iR_x)^2$ is identity on quaternion subalgebra $\langle i, x\rangle$ and minus identity on perpendicular 4-space. Set $M$ is topologically $\mathbb CP^2$ and generates $SU_3$.