There is a property for the 3-j symbols as
$$ \sum_{m_1 m_2 m_3} \begin{pmatrix} j_1 & j_2 & j_3 \\ m_1 & m_2 & m_3 \end{pmatrix}^2 = \Delta (j_1, j_2, j_3), $$ where $ \Delta (j_1, j_2, j_3)$ is the triangular condition i.e. if it is satisfied, it equals to the unit otherwise is zero.
I have checked this property numerically and it's working well. However, I need to prove it analytically. How can I figure it out? Thanks
I fear you have misconfigured something. The standard orthogonality condition reflecting the unitarity of the 3-j symbols, as checked in WP, $m_3=m_3'=-m_1-m_2$, $j_3'=j_3$ is $$ (2j_3+1)\sum_{m_1, m_2} \begin{pmatrix} j_1 & j_2 & j_3 \\ m_1 & m_2 & m_3 \end{pmatrix}^2 = \Delta (j_1, j_2, j_3), $$ so, expressible in terms of the CG coefficients, the orthogonality condition of the CG reduction matrix, $$ \sum_{m_1, m_2} \langle j_3 \, m_3 | j_1 \, m_1 \, j_2 \, m_2 \rangle \langle j_1 \, m_1 \, j_2 \, m_2 | j_3 \, m_3 \rangle = \Delta (j_1, j_2, j_3). $$ The triangular condition is baked into the definition of CG coefficients, so it is implied when one writes this "second" orthogonality condition out, as physicists learn in elementary QM the conditions of the reduction of $j_1\otimes j_2$ and apply them as they breathe.