Explanation

(a)
Taking $x$ common from $R_{2}$and$\ x(x - 1)$common from $R_{3}$,we get
$f(x) = x^{2}(x - 1)\left| \begin{matrix}
1 & x & (x + 1) \\
2 & (x - 1) & (x + 1) \\
3 & (x - 2) & (x + 1)
\end{matrix} \right|$
$\Rightarrow f(x) = x^{2}(x - 1)(x + 1)\left| \begin{matrix}
1 & x & 1 \\
2 & x - 1 & 1 \\
3 & x - 2 & 1
\end{matrix} \right|$
$= x^{2}\left( x^{2} - 1 \right)\left| \begin{matrix}
1 & x & 1 \\
1 & - 1 & 0 \\
2 & - 2 & 0
\end{matrix} \right|\ \ \ \ \left\lbrack \begin{array}{r}
R_{2} \rightarrow R_{2} - R_{1} \\
R_{3} \rightarrow R_{3} - R_{1}
\end{array} \right\rbrack$
$\Rightarrow f(x) = x^{2}\left( x^{2} - 1 \right)( - 2 + 2) = 0$
$\Rightarrow \ \ \ f(x) = 0$for all $x$
$\therefore\ \ \ \ f(11) = 0$