The Intersection

$\begin{cases} x + ky + 3z = 0 & ...(1) \\ 3x + ky - 2z = 0 & ...(2) \\ 2x+4y-3z = 0 & ...(3) \end{cases}$

$\begin{aligned} (2)-(1): 2x- 5z & = 0 \\ \implies z & = \frac 25 x \\ (3): 2x + 4y - \frac 65x & = 0 \\ \implies y & = - \frac 15 x \end{aligned}$

Let us check if there is a $k$ that satisfies the system of equations.

$\begin{cases} (1): x - \dfrac 15 kx + \dfrac 65 x = 0 & \implies k = 11 \\ (2): 3x - \dfrac 15 kx - \dfrac 45 x = 0 & \implies k = 11 \end{cases}$

Therefore there is a unique $k$ satisfying the system of equations and $\dfrac {xz}{y^2} = \dfrac {x\left(\frac 25x\right)}{\left(-\frac 15x\right)^2} = \boxed{10}$ .

For the matrix equation $M\mathbf{r} = \mathbf{0}$ where $M \; = \; \left(\begin{array}{ccc} 1 & k & 3 \\ 3 & k & -2 \\ 2 & 4 & -3 \end{array}\right) \hspace{2cm} \mathbf{r} \; = \; \left(\begin{array}{c} x \\ y \\ z \end{array}\right)$ to have a solution with nonzero $\mathbf{r}$ , we must have $\mathrm{det}\,M = 0$ . But $\mathrm{det}\,M = 44 - 4k$ , so we deduce that $k = 11$ . It is easy now to show that $\mathbf{r} \; = \; \left(\begin{array}{c} 5\alpha \\ -\alpha \\ 2\alpha \end{array} \right)$ for some $\alpha \neq 0$ , making the answer $\boxed{10}$ .

Write the system in the standard form $A x' = b$ . We notice that the $b$ vector is full of zeroes. Therefore, if matrix $A$ is invertible (non-singular), we will get a trivial solution in which $(x,y,z) = (0,0,0)$ . So if there is to be a non-zero solution, the system matrix must be singular (its determinant must be zero). In order for the determinant to be zero, the constant $k$ must equal $11$ . Then solving the system by whatever means yields many possible solutions, but the quantity $\large{\frac{x z}{y^2}}$ is always equal to $10$ . I determined that just by solving it many times using a numerical routine. I'm sure somebody will find a clever way of manipulating the equations to arrive at the answer.

for the equations to have non zero solution, the following determinant should be zero:

i.e. ||1||k||3|| ||3||k||-2|| =0 ||2||4||-3||

(read above as a determinant, not as a table)

this means, on solving the determinant, k=11

now suppose z=a and solve the given equations putting the value of k, and find x and y in terms of a. Then put in (xz)/(y^2), and find the answer.

EDIT: My doubt has been cleared. I was just equating the third plane's equation with the equation of family of planes though intersection of the other other 2 planes, instead of making them proportional. Wholehearted thanks to all of you who replied :)