Let $\left|\begin{array}{ccc}x & 2 & x \\ x^{2} & x & 6 \\ x & x & 6\end{array}\right|=a x^{4}+b x^{3}+c x^{2}+d x+e$
Then, the value of $5 a+4 b+3 c+2 d+e$ is equal to
(a) 0
(b) $-16$
(c) 16
(d) none of these
(d) none of these
$\Delta=\mid \begin{array}{lll}x & 2 & x\end{array}$
$x^{2} \quad x \quad 6$
$\begin{array}{lll}x & x & 6\end{array}$
$=x \mid x \quad 6$
$\begin{array}{ll}x & 6\left|-x^{2}\right| 2 x \\ x & 6|+x| 2 x\end{array}$
$\begin{array}{ll}x & 6\end{array}$ [Expanding along $C_{1}$ ]
$=0-x^{2}\left(12-x^{2}\right)+x\left(12-x^{2}\right)$
$=x^{4}-12 x^{2}+12 x-x^{3}$
$\Delta=a x^{4}+b x^{3}+c x^{2}+d x+e$ [Given]
$\Rightarrow x^{4}-12 x^{2}+12 x-x^{3}=a x^{4}+b x^{3}+c x^{2}+d x+e$
$\Rightarrow a=1, b=-1, c=-12, d=12, e=0$
Thus,
$5 a+4 b+3 c+2 d+e=5-4-36+24+0=-11$