A root of the equation left| begin{array}{ccc | vedclass

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(C) Given the determinant equation: $\left| \begin{array}{ccc} 3 - x & -6 & 3 \ -6 & 3 - x & 3 \ 3 & 3 & -6 - x \end{array} ight| = 0$. Applying t

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$0$

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(C) Given the determinant equation: $\left| \begin{array}{ccc} 3 - x & -6 & 3 \ -6 & 3 - x & 3 \ 3 & 3 & -6 - x \end{array} ight| = 0$. Applying the column operation $C_1 	o C_1 + C_2 + C_3$: $\left| \begin{array}{ccc} 3 - x - 6 + 3 & -6 & 3 \ -6 + 3 - x + 3 & 3 - x & 3 \ 3 + 3 - 6 - x & 3 & -6 - x \end{array} ight| = 0$ $\Rightarrow \left| \begin{array}{ccc} -x & -6 & 3 \ -x & 3 - x & 3 \ -x & 3 & -6 - x \end{array} ight| = 0$ Taking $-x$ common from $C_1$: $-x \left| \begin{array}{ccc} 1 & -6 & 3 \ 1 & 3 - x & 3 \ 1 & 3 & -6 - x \end{array} ight| = 0$ Applying $R_2 	o R_2 - R_1$ and $R_3 	o R_3 - R_1$: $-x \left| \begin{array}{ccc} 1 & -6 & 3 \ 0 & 9 - x & 0 \ 0 & 9 & -9 - x \end{array} ight| = 0$ Expanding along $C_1$: $-x [1 \cdot ((9 - x)(-9 - x) - 0)] = 0$ $-x [-(9 - x)(9 + x)] = 0$ $x(9 - x)(9 + x) = 0$ Thus,the roots are $x = 0, 9, -9$. Comparing with the options,$0$ is a root.

$A$ root of the equation $\left| \begin{array}{ccc} 3 - x & -6 & 3 \\ -6 & 3 - x & 3 \\ 3 & 3 & -6 - x \end{array} \right| = 0$ is

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