b
(b) $\left| {\,\begin{array}{*{20}{c}}{1 + x}&1&1\\1&{1 + x}&1\\1&1&{1 + x}\end{array}\,} \right|\, = \,0$
$ \Rightarrow $$\left| {\,\begin{array}{*{20}{c}}
{3 + x}&0&1\\
{3 + x}&x&1\\
{3 + x}&{ - x}&{1 + x}
\end{array}\,} \right|\, = \,0$, $\left( \begin{array}{l}{C_1} \to {C_1} + {C_2} + {C_3}\\{C_2} \to {C_2} - {C_3}\end{array} \right)$
$ \Rightarrow $ $(x + 3)\,\left| {\,\begin{array}{*{20}{c}}1&0&1\\1&x&1\\1&{ - x}&{1 + x}\end{array}\,} \right|\, = 0$
$ \Rightarrow $ $(x + 3)\,\left| {\,\begin{array}{*{20}{c}}1&0&1\\0&x&0\\0&{ - x}&x\end{array}\,} \right|\, = 0$, $\left( \begin{array}{l}{R_2} \to {R_2} - {R_1}\\{R_3} \to {R_3} - {R_1}\end{array} \right)$
$ \Rightarrow $ $(x + 3){x^2} = 0 \Rightarrow x = 0,\,0,\, - 3$.
Trick : Obviously the equation is of degree three, therefore it must have three solutions.
So check for option $ (b).$