The system of equation $2 x + 3 y - z = 0, 3 x + 2 y + k z = 0$ and $4 x + y + z = 0$ have a set of non-zero integral solution, then the least positive value of $z$ is:

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Solution

$2 x + 3 y - z = 0 3 x + 2 y + k z = 0 4 x + y + z = 0 ⎡ ⎢ ⎣ \begin{matrix} 000 \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 2 & 3 & - 1 3 & 2 & k 4 & 1 & 1 \end{matrix} ⎤ ⎥ ⎦ . . . . . . [CRAMER'S RULE] Δ = 2 (2 - K) - 3 (3 - 4 K) - 1 (3 - 8) = 4 - 2 K - 9 + 12 K + 5 = 10 K$
Now, $x = \frac{△_{1}}{△}, y = \frac{△_{2}}{△}, z = \frac{△_{3}}{△}$
Clearly, $△_{1} = △_{2} = △_{3} = 0$
$\Rightarrow K \neq 0$
$\Rightarrow x = y = z = 0$ , which is not possible.

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