= $\left| {\,\begin{array}{*{20}{c}}
{ - 1}&{ - 1}&x\\
{ - 2}&{ - 1}&x\\
{ - 3}&{ - 1}&{x + 2}
\end{array}\,} \right|,$  ,     by $\begin{array}{l}{C_2} \to {C_2} - {C_1}\\{C_3} \to {C_3} + 4{C_1}\end{array}$

$ = - ( - x - 2 + x) + 1\,.\,( - 2x - 4 + 3x) + x(2 - 3)$

= $2 + x - 4 - x = - 2$.

Trick : Put $x=1$. Then $\left| {\,\begin{array}{*{20}{c}}2&3&5\\4&6&9\\8&{11}&{15}\end{array}\,} \right| = - 2$

Note : Since there is a option “None of these”, therefore we should check for one more different value of $x$. Put $x = - 1$.

$\left| {\,\begin{array}{*{20}{c}}0&1&3\\2&4&7\\6&9&{13}\end{array}\,} \right| = - 1(26 - 42) + 3(18 - 24) = - 2$

Therefore answer is  $ (b).$