Let us consider three mutually perpendicular planes
$l_ix+m_iy+n_iz=p_i$ for $i=1,2,3$, where $l_i,m_i,n_i$ are direction cosines of the normals to the planes. Since these three planes are mutually perpendicular, we have
$l_il_j+m_im_j+n_in_j=0$ for $i\neq j$. Also we can say that $l_i^2+m_i^2+n_i^2=1$ for $i=1,2,3$.
How can I get the following relation from the above conditions:
$$l_1m_1+l_2m_2+l_3m_3=0$$ etc.
The given condition implies $$A^TA=\left(\begin{matrix} l_1&m_1&n_1\\l_2&m_2&n_2\\ l_3&m_3&n_3 \end{matrix}\right)\left(\begin{matrix} l_1&l_2&l_3\\m_1&m_2&m_3\\ n_1&n_2&n_3 \end{matrix}\right)=I.$$ This implies $A^T=A^{-1}$ and hence $$ AA^T=I. $$ Then try to compute $(1,2)$-th entry of $AA^T$ (and others, too.)