Mathematics - Determinant Question with Solution | TestHub

$\left|\begin{array}{ccc}1& k& 3\\ 3& k& -2\\ 2& 4& -3\end{array}\right|=0$

$\Rightarrow \left(-3k+8\right)-3\left(-3k-12\right)+2\left(-5k\right)=0$

$\Rightarrow -4k+44=0$$\Rightarrow$$k=11$

$\Rightarrow x+11y+3z=0 ...\left(1\right)$

$\Rightarrow 3x+11y-2z=0 ...\left(2\right)$

$\Rightarrow 2x+4y-3z=0 ...\left(3\right)$

On solving equations$\left(1\right) \& \left(2\right)$, we get

$2x-5z=0$

$\Rightarrow x=\frac{5z}{2}$

Put it in equation$\left(3\right)$, we get

$2z+4y=0$

$\Rightarrow z=-2y$

$\therefore \frac{xz}{y^2}=\frac{\left(-2y\right)\times \left(-5y\right)}{y^2}=10$