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(D) Expanding the determinant along the first row:$(x - 1) \cdot \left| \begin{array}{cc} x - 3 & 4 \ 5 & 6 \end{array} ight| - 3 \cdot \left| \beg

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$1$

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(D) Expanding the determinant along the first row:$(x - 1) \cdot \left| \begin{array}{cc} x - 3 & 4 \ 5 & 6 \end{array} ight| - 3 \cdot \left| \begin{array}{cc} 2 & 4 \ 3 & 6 \end{array} ight| + 0 \cdot \left| \begin{array}{cc} 2 & x - 3 \ 3 & 5 \end{array} ight| = 0$$(x - 1) \cdot [6(x - 3) - 20] - 3 \cdot [12 - 12] + 0 = 0$$(x - 1) \cdot [6x - 18 - 20] - 3 \cdot [0] = 0$$(x - 1)(6x - 38) = 0$$2(x - 1)(3x - 19) = 0$Thus,$x - 1 = 0$ or $3x - 19 = 0$.Therefore,$x = 1$ or $x = \frac{19}{3}$.Comparing with the given options,$x = 1$ is the correct choice.

If $\left| \begin{array}{ccc} x - 1 & 3 & 0 \\ 2 & x - 3 & 4 \\ 3 & 5 & 6 \end{array} \right| = 0$,then $x =$

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