If $\begin{bmatrix}a & b & c\\m & n & p\\x & y & z\end{bmatrix}=k, $ then $\begin{bmatrix}6a & 2b & 2c\\3m & n & p\\3x & y & z\end{bmatrix}=$

$6k$

The correct answer is option (4) : $6k$

Taking 3 Common from $C_1$ and 2 from $R_1$

$\begin{bmatrix}6a & 2b & 2c\\3m & n & p\\3x & y & z\end{bmatrix}=3×2\begin{bmatrix}a & b & c\\m & n & p\\x & y & z\end{bmatrix}=6k$