If the system of linear equations.
`x+ky+3z=0`
`3x+ky-2z=0`
`2x+4y-3z=0`
has a non-zero solution (x,y,z) then .`(xz)/(y^(2))` is equal to

Options

A. `-10`
B. 10
C. `-30`
D. 30

Explanation

We have,
x + ky +3z = 0, 3x + ky-2z =0, 2x + 4y-3z =0
System of equation has non-zero solution, if
`|{:(1, k, 3), (3, k, -2), (2, 4, -3):}|= 0`
`rArr (-3k + 8)-k(-9+4) +3(12-2k) =0`
`rArr -3k + 8 + 9k-4k + 36-6k = 0`
`rArr -4k + 44 = 0 rArr k = 11`
Let `z = lambda` then we get
` x + 11y+ 3lambda =0 " "...(i)`
`3x + 11y -2 lambda =0 " "...(ii)`
`"and " 2x + 4y-3lambda =0" "...(iii)`
Solving Eqs. (i) and (ii), we get
`x = (5lambda)/(2), y = (-lambda)/(2), z = lambda rArr (xz)/(y^(2)) = (5lambda^(2))/(2 xx (-(lambda)/(2))^(2)) = 10`