JEE Main | 2018Math miscellaneousHard

Question

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 xzy2 is equal to :

Options

A.

10

B.

- 30

C.

30 

D.

-10

Solution

For non zero solution 1k33k-224-3 = 0 
⇒ k = 11
Now equations
x + 11y + 3z = 0      .....(1)
3x + 11y - 2z = 0     .....(2)
2x + 4y - 3z = 0      .....(3)
on equation (1) + (3) we get 3x + 15y = 0
⇒ x = - 5y
Now put x = - 5y in equation (1)
we get - 5y + 11y + 3z = 0
⇒ z = - 2y
xzy2=(-5y)(-2y)y2=10

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