If one of the roots of the equation 
$\left|\begin{array}{ccc}7& 6& {\mathrm{x}}^2-13\\ 2& {\mathrm{x}}^2-13& 2\\ {\mathrm{x}}^2-13& 3& 7\end{array}\right|=0$ is 2,  then the square of the sum of the remaining five roots is:

Let ${\mathrm{x}}^2-13=\mathrm{t}$

Expnad the determinant, we get

${\mathrm{t}}^3-67\mathrm{t}+126=0$
$\Rightarrow \mathrm{t}=-9,7,2$

Hence, the roots are $\mathrm{x}=\pm 2,\pm \sqrt{20},\pm \sqrt{15}$

Sum of the other 5 roots is $-2$, square of it is $4$