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(A) Given the determinant equation $\left|\begin{array}{ccc}x & 2 & 2 \ 2 & x & 2 \ 2 & 2 & x\end{array}ight|=0$. Applying $C_1 ightarrow C_1-C_

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$6$

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(A) Given the determinant equation $\left|\begin{array}{ccc}x & 2 & 2 \ 2 & x & 2 \ 2 & 2 & x\end{array}ight|=0$. Applying $C_1 ightarrow C_1-C_2$ and $C_2 ightarrow C_2-C_3$,we get: $\left|\begin{array}{ccc}x-2 & 0 & 2 \ 2-x & x-2 & 2 \ 0 & 2-x & x\end{array}ight|=0$. Taking $(x-2)$ common from $C_1$ and $C_2$: $(x-2)^2 \left|\begin{array}{ccc}1 & 0 & 2 \ -1 & 1 & 2 \ 0 & -1 & x\end{array}ight|=0$. Applying $R_2 ightarrow R_2+R_1$: $(x-2)^2 \left|\begin{array}{ccc}1 & 0 & 2 \ 0 & 1 & 4 \ 0 & -1 & x\end{array}ight|=0$. Expanding along $C_1$: $(x-2)^2 [1(x+4)-0+0]=0 \Rightarrow (x-2)^2(x+4)=0$. The roots are $x = 2, 2, -4$. Given $\min(\alpha, \beta, \gamma) = \alpha$,we have $\alpha = -4$,$\beta = 2$,and $\gamma = 2$. Calculating the value: $2\alpha + 3\beta + 4\gamma = 2(-4) + 3(2) + 4(2) = -8 + 6 + 8 = 6$.

If $\alpha, \beta, \gamma$ are the roots of the equation $\left|\begin{array}{ccc}x & 2 & 2 \\ 2 & x & 2 \\ 2 & 2 & x\end{array}\right|=0$ and $\min (\alpha, \beta, \gamma)=\alpha$,then $2 \alpha+3 \beta+4 \gamma=$

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