The following recursion relation
$$
C(m_2+1,m_3-1)\begin{pmatrix}l_1 &l_2 &l_3\\m_1&m_2+1&m_3-1\end{pmatrix}+
D(m_2,m_3)\begin{pmatrix}l_1 &l_2 &l_3\\m_1&m_2&m_3\end{pmatrix}+$$ $$
C(m_2,m_3)\begin{pmatrix}l_1 &l_2 &l_3\\m_1&m_2-1&m_3+1\end{pmatrix}=0,
\tag{1}$$
where $$C(m_2,m_3)=\sqrt{(l_2-m_2+1)(l_2+m_2)(l_3+m_3+1)(l_3-m_3)},\tag{2}$$ and
$$D(m_2,m_3)=l_2(l_2+1)+l_3(l_3+1)-l_1(l_1+1)+2m_2m_3,\tag{3}$$
can be found (formula 9) in http://scitation.aip.org/content/aip/journal/jmp/16/10/10.1063/1.522426?ver=pdfcov (Exact recursive evaluation of 3j‐ and 6j‐coefficients for quantum‐mechanical coupling of angular momenta, by Klaus Schulten and Roy G. Gordon). For $m_1=m_2=m_3=0$, we get from (1):
$$
C(1)\begin{pmatrix}l_1 &l_2 &l_3\\0&1&-1\end{pmatrix}+
D(0)\begin{pmatrix}l_1 &l_2 &l_3\\0&0&0\end{pmatrix}+
C(0)\begin{pmatrix}l_1 &l_2 &l_3\\0&-1&1\end{pmatrix}=0.
\tag{4}$$
Using the symmetry relation
$$\begin{pmatrix}l_1 &l_2 &l_3\\0&-1&1\end{pmatrix}=(-1)^{l_1+l_2+l_3}
\begin{pmatrix}l_1 &l_2 &l_3\\0&1&-1\end{pmatrix},$$
(4) can be rewritten as follows
$$[C(1)+(-1)^{l_1+l_2+l_3}C(0)]\begin{pmatrix}l_1 &l_2 &l_3\\0&1&-1\end{pmatrix}+D(0)\begin{pmatrix}l_1 &l_2 &l_3\\0&0&0\end{pmatrix}=0.\tag{5}$$
Here, according to (2) and (3), $$C(0)=\sqrt{l_2l_3(l_2+1)(l_3+1)},\;\;
C(1)=C(0),$$ and $$D(0)=l_2(l_2+1)+l_3(l_3+1)-l_1(l_1+1).$$
Now let us assume $m_1=0,m_2=1,m_3=-1$ in (1). We get
$$
\tilde C(2)\begin{pmatrix}l_1 &l_2 &l_3\\0&2&-2\end{pmatrix}+
\tilde D(1)\begin{pmatrix}l_1 &l_2 &l_3\\0&1&-1\end{pmatrix}+
\tilde C(1)\begin{pmatrix}l_1 &l_2 &l_3\\0&0&0\end{pmatrix}=0.
$$
so that
$$\begin{pmatrix}l_1 &l_2 &l_3\\0&1&-1\end{pmatrix}=-\frac{\tilde C(2)}
{\tilde D(1)}\begin{pmatrix}l_1 &l_2 &l_3\\0&2&-2\end{pmatrix}
-\frac{\tilde C(1)}
{\tilde D(1)}\begin{pmatrix}l_1 &l_2 &l_3\\0&0&0\end{pmatrix}.$$
Substituting this into (5), we get after a little algebra
$$\frac{\begin{pmatrix}l_1 &l_2 &l_3\\0&2&-2\end{pmatrix}}
{\begin{pmatrix}l_1 &l_2 &l_3\\0&0&0\end{pmatrix}}=\frac{1}{\tilde C(2)}\left[\frac{D(0)\tilde D(1)}{C(1)+(-1)^{l_1+l_2+l3}C(0)}-\tilde C(1)\right ].$$ It remains to take into account that $l_1+l_2+l_3$ is even and calculate from (2) and (3) (for $m_3=-1$)
$$\tilde C(1)=C(1),\;\;\tilde C(2)=\sqrt{(l_2-1)(l_2+2)(l_3-1)(l_3+2)},\;\;\;
\tilde D(1)=D(0)-2.$$ Finally,
$$\frac{\begin{pmatrix}l_1 &l_2 &l_3\\0&2&-2\end{pmatrix}}
{\begin{pmatrix}l_1 &l_2 &l_3\\0&0&0\end{pmatrix}}=\frac{1}{\tilde C(2)}\left[\frac{D(0)[D(0)-2]}{2C(0)}-C(0)\right ].$$
