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The system of equations
$x + k y + 3 z = 0$ ,
$3 x + k y - 2 z = 0$ ,
$2 x + 3 y - 4 z = 0$ ,
has a non-trivial solution then $k =$

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-33

$- \frac{33}{2}$

$\frac{33}{2}$

answer is C.

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Detailed Solution

$|\begin{matrix} 1 & k & 3 \\ 3 & k & - 2 \\ 2 & 3 & - 4 \end{matrix}| = 0$

$(- 4 k + 6) - k (- 12 + 4) + 3 (9 - 2 k) = 0$

$- 2 k + 33 = 0$

$k = \frac{33}{2}$

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