Explanation

(a)
We have, $\left| \begin{matrix}
x & 3 & 7 \\
2 & x & 2 \\
7 & 6 & x
\end{matrix} \right| = 0$
$\Rightarrow (x + 9)\left| \begin{matrix}
1 & 1 & 1 \\
2 & x & 2 \\
7 & 6 & x
\end{matrix} \right| = 0\ \ (R_{1} \rightarrow R_{1} + R_{2} + R_{3})$
$\Rightarrow (x + 9)\left\{ 1\left( x^{2} - 12 \right) - 1(2x - 14) + 1(12 - 7x) \right\} = 0$
$\Rightarrow (x + 9)\left( x^{2} - 9x + 14 \right) = 0$
$\Rightarrow \ (x + 9)(x - 2)(x - 7) = 0$
$\therefore$ The other two roots are 2 and 7.