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(A) Given determinant is $\left| \begin{array}{ccc} x+1 & 1 & 1 \ 2 & x+2 & 2 \ 3 & 3 & x+3 \end{array} ight| = 0$. Applying $C_1 	o C_1 - C_2$ a

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$0, -6$

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(A) Given determinant is $\left| \begin{array}{ccc} x+1 & 1 & 1 \ 2 & x+2 & 2 \ 3 & 3 & x+3 \end{array} ight| = 0$. Applying $C_1 	o C_1 - C_2$ and $C_2 	o C_2 - C_3$: $\left| \begin{array}{ccc} x & 0 & 1 \ -x & x & 2 \ 0 & -x & x+3 \end{array} ight| = 0$. Taking $x$ common from $C_1$ and $C_2$: $x^2 \left| \begin{array}{ccc} 1 & 0 & 1 \ -1 & 1 & 2 \ 0 & -1 & x+3 \end{array} ight| = 0$. Expanding along $R_1$: $x^2 [1(x+3+2) - 0 + 1(1-0)] = 0$. $x^2 (x+5+1) = 0$. $x^2 (x+6) = 0$. Thus,$x = 0$ or $x = -6$.

If $\left| \begin{array}{ccc} x+1 & 1 & 1 \\ 2 & x+2 & 2 \\ 3 & 3 & x+3 \end{array} \right| = 0$,then $x$ is

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