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If ∣2+xyzx2+yzxy2+z∣=0\begin{vmatrix} 2+x & y & z \\ x & 2+y & z \\ x & y & 2+z \end{vmatrix} = 0, then the value of x+y+zx+y+z is

Solution

✅ Correct Option: 2

Apply C1→C1+C2+C3C_1\to C_1+C_2+C_3; every entry of the first column becomes 2+x+y+z2+x+y+z, which factors out. Row operations R2−R1R_2-R_1 and R3−R1R_3-R_1 reduce the remaining determinant to 4. So 4(2+x+y+z)=04(2+x+y+z)=0, giving x+y+z=−2x+y+z=-2.

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