Solution:

Option(C)
To find: Value of $\mathrm{x}$
We have, $\left|\begin{array}{ccc}5 & 3 & -1 \\ -7 & x & 2 \\ 9 & 6 & -2\end{array}\right|=0$
Applying $R_{1} \rightarrow 2 R_{1}$
$\Rightarrow\left|\begin{array}{ccc}
10 & 6 & -2 \\
-7 & x & 2 \\
9 & 6 & -2
\end{array}\right|=0$
Applying $R_{1} \rightarrow R_{1}-R_{3}$
$\Rightarrow\left|\begin{array}{ccc}
1 & 0 & 0 \\
-7 & x & 2 \\
9 & 6 & -2
\end{array}\right|=0$
Expanding along $R_{1}$
$\begin{array}{l}
\Rightarrow[1\{(x)(-2)-(6)(2)\}]=0 \\
\Rightarrow[1\{-2 x-12\}]=0 \\
\Rightarrow-2 x-12=0 \\
\Rightarrow-2 x=12 \\
\Rightarrow x=-6
\end{array}$