Let $A=\left|\begin{array}{lll}x & 2 & x \\ x^2 & x & 6 \\ x & x & 6\end{array}\right| = p x^4+q x^3+r x^2+s x+t$, then the value of $5 p+4 q+3 r+2 s+t$ is equal to:

-11

The correct answer is Option (3) → -11

$\left|\begin{array}{lll}x & 2 & x \\ x^2 & x & 6 \\ x & x & 6\end{array}\right|=x\begin{vmatrix}x&6\\x&6\end{vmatrix}-2\begin{vmatrix}x^2&6\\x&6\end{vmatrix}+x\begin{vmatrix}x^2&x\\x&x\end{vmatrix}$

so $x^4-x^3+12x^2+12x=9x^4+qx^3+5x^2+5x+t$

$p=1,q=-1,r=-12,s=12,t=0$

$5p+4q+3r+2s+t$

$=5-4-36+24+0$

$=1-12=-11$