If $x, y$ and $z$ are real number such that $x + y + z = 0$, then value of $\begin{vmatrix}3x&-x+y&-x+z\\x-y&3y&z-y\\x-z&y-z&3z\end{vmatrix}$ is

0

The correct answer is Option (3) → 0

Given $x+y+z=0$ and matrix $M=\begin{vmatrix} 3x & -x+y & -x+z\\[4pt] x-y & 3y & z-y\\[4pt] x-z & y-z & 3z \end{vmatrix}$

Sum of entries in each row:

Row-1: $3x+(-x+y)+(-x+z)=x+(y+z)=x+(-x)=0$

Row-2: $(x-y)+3y+(z-y)=x+y+z=0$

Row-3: $(x-z)+(y-z)+3z=x+y+z=0$

The value of the determinant is $0$.