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Q)

# If $\begin{vmatrix}x+\alpha&\beta&\gamma\\\alpha&x+\beta&\gamma\\\alpha&\beta&x+\gamma\end{vmatrix}=0$ then $4x$ is equal to

$\begin{array}{1 1}(A)\;0,-(\alpha+\beta+\gamma)&(B)\;0,(\alpha+\beta+\gamma)\\(C)\;1,(\alpha+\beta+\gamma)&(D)\;0,(\alpha^2+\beta^2+\gamma^2)\end{array}$

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A)
$\begin{vmatrix}x+\alpha&\beta&\gamma\\\alpha&x+\beta&\gamma\\\alpha&\beta&x+\gamma\end{vmatrix}=0$
Apply $C_1=C_1+C_2+C_3$
$\begin{vmatrix}x+\alpha+\beta+\gamma&\beta&\gamma\\x+\alpha+\beta+\gamma&x+\beta&\gamma\\x+\alpha+\beta+\gamma&\beta&x+\gamma\end{vmatrix}=0$
$(x+\alpha+\beta+\gamma)\begin{vmatrix}1&\beta&\gamma\\1&x+\beta&\gamma\\1&\beta&x+\gamma\end{vmatrix}=0$
$(x+\alpha+\beta+\gamma)[(x^2+x\gamma+\beta\gamma+\gamma\beta-\gamma\beta)+x\beta+\beta\gamma-\gamma\beta+\beta\gamma-\beta x-\beta \gamma]$
$(x+\alpha+\beta+\gamma)[x^2+x\gamma+\beta\gamma+\gamma\beta-\gamma\beta+x\beta+\beta\gamma-\gamma\beta+\beta\gamma-\beta x-\beta \gamma]$
$(x+\alpha+\beta+\gamma)[x^2]=0$
$\Rightarrow x=0,-(\alpha+\beta+\gamma)$
Hence (A) is the correct answer.