If $\begin{vmatrix}x+4 & 2x & 2x\\2x&x+4&2x\\2x & 2x & x+4\end{vmatrix}=k(x-4)^2 $, then value of k is :

$5x+4$

The correct answer is Option (3) → $5x+4$

$\begin{vmatrix}x+4 & 2x & 2x\\2x&x+4&2x\\2x & 2x & x+4\end{vmatrix}$

$R_1→R_1+R_2+R_3$

$\begin{vmatrix}5x+4 & 5x+4 & 5x+4\\2x&x+4&2x\\2x & 2x & x+4\end{vmatrix}$

$⇒(5x+4)\begin{vmatrix}1 & 1 & 1\\2x&x+4&2x\\2x & 2x & x+4\end{vmatrix}$

$C_2→C_2-C_1,C_3→C_3-C_1$

$(5x+4)\begin{vmatrix}1 & 0 & 0\\2x&-x+4&0\\2x & 0 & -x+4\end{vmatrix}$

$=(5x+4)(x-4)^2$

so $k=5x+4$