Given $\left|\begin{array}{ccc}x & 3 & 7 \\ 2 & x & 2 \\ 7 & 6 & x\end{array}\right|=0$
$\begin{array}{l}
\Rightarrow x\left[x^{2}-12\right]-3[2 x-14]+7[12-7 x]=0 \\
\Rightarrow x^{3}-67 x+126=0
\end{array}$
But given $(x=9)$ is a root of given determinant.
$\therefore(\mathrm{x}+9)$ is a factor
$\begin{array}{l}
\Rightarrow \mathrm{x}^{3}+9 \mathrm{x}^{2}-9 \mathrm{x}^{2}-81 \mathrm{x}+14 \mathrm{x}+126=0 \\
\Rightarrow \mathrm{x}^{2}(\mathrm{x}+9)-9 \mathrm{x}(\mathrm{x}+9)+14(\mathrm{x}+9)=0
\end{array}$
$\Rightarrow(x+9)\left(x^{2}-9 x+14\right)=0$
$\Rightarrow(x+9)\left(x^{2}-7 x-2 x+14\right)=0$
$\Rightarrow(x+9)(x-7)(x-2)=0$
$\Rightarrow x=-9,7,2$