Answer:
Option B
Explanation:
We have, x+ky+3z=0; 3x+ky-2z=0;2x+4y-3z=0
System of the equation has non-zero solution, If
$\begin{bmatrix}1 & k & 3 \\3 & k & -2 \\2 & 4 & -3 \end{bmatrix}=0$
$\Rightarrow$ (-3k+8)-k(-9+4)+3(12-2k)=0
$\Rightarrow$ -3k+8+9k-4k+36-6k=0
$\Rightarrow$ -4k+44=0 $\Rightarrow$ k=11
Let z=λ , then we get
x+11y+3λ =0 ...... (i)
3x+11y-2λ=0 .......(ii)
and 2x+4y-3λ =0 ........(iii)
Solving Eqs. (i) and (ii) , we get
$x=\frac{5\lambda}{2},y=\frac{-\lambda}{2}, z=\lambda$
$\Rightarrow$ $\frac{xz}{y^{2}}=\frac{5\lambda^{2}}{2\times (-\frac{\lambda}{2})^{2}}=10$
