(C) To find the value of $x$,we expand the determinant along the first row:$\left| \begin{array}{ccc} 1 & 2 & 3 \\ 2 & x & 3 \\ 3 & 4 & 5 \end{array}

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(C) To find the value of $x$,we expand the determinant along the first row:$\left| \begin{array}{ccc} 1 & 2 & 3 \\ 2 & x & 3 \\ 3 & 4 & 5 \end{array} \right| = 1(5x - 12) - 2(10 - 9) + 3(8 - 3x) = 0$Expanding the terms:$(5x - 12) - 2(1) + 3(8 - 3x) = 0$$5x - 12 - 2 + 24 - 9x = 0$Combining like terms:$-4x + 10 = 0$$-4x = -10$$x = \frac{-10}{-4} = \frac{5}{2}$Thus,the correct value of $x$ is $\frac{5}{2}$.

Class 12 Mathematics 3-4. Matrices and Determinants Expansion of determinants, Solution of equation in the form of determinants and area of triangle and Equation of Line

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

A
$-5/2$
B
$-2/5$
C
$5/2$
D
$2/5$

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Expansion of determinants, Solution of equation in the form of determinants and area of triangle and Equation of Line

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If $q_1, q_2, q_3$ are roots of the equation $x^3 + 64 = 0$,then the value of $\left| \begin{array}{ccc} q_1 & q_2 & q_3 \\ q_2 & q_3 & q_1 \\ q_3 & q_1 & q_2 \end{array} \right|$ is

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

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