If $\left|\begin{array}{ccc}y+z & x & x \\ y & z+x & y \\ z & z & x+y\end{array}\right|=k(x y z)$, then k is equal to

4

The correct answer is Option (1) → 4

$\left|\begin{array}{ccc}y+z & x & x \\ y & z+x & y \\ z & z & x+y\end{array}\right|=k(x y z)$

Putting, $x=y=z=1$

$\left|\begin{array}{ccc}2& 1 &1 \\ 1 &2 &1 \\ 1 & 1 & 2\end{array}\right|=k$

$⇒2(4-1)-1(2-1)+1(1-2)=k$

$⇒6-2=k$

$⇒k=4$