Find all values of `lambda` for which the `(lambda-1)x+(3lambda+1)y+2lambdaz=0, (lambda-1)x+(4lambda-2)y+(lambda+3)z=0, 2x+(3lambda+1)y+3(lambda-1)z=0` possess non-trivial solution and find the ratios `x:y:z,` where `lambda` has the smallest of these value.

Options

A. `3 : 2 :1`
B. `3 : 3 : 2`
C. `1 : 3 : 1`
D. `1 : 1 : 1`

Answer

D

Explanation

`(d)` `|{:(lambda-1,3lamda+1,2lambda),(lambda-1,4lambda-2,lambda+3),(2,3lambda+1,3(lambda-1)):}|=0`
`implieslambda=0` or `3`
If `lambda=0`, the equations become
`-x+y=0`,
`-x-2y+3z=0` and
`2x+y-3z=0`,
`:. (x)/(6-3)=(y)/(6-3)=(z)/(-1+4)`