Correct Answer - A
Now, [a b c] `=| (lambda, 2 , -3),(2,lambda,-1),(1,2,1)|`
On applying `C_(1) to C_(1)-C_(3)`, we get
`=|(lambda+3, 2 , -3),(3, lambda, -1),(0,2,1)|`
`(lambda+3)(lambda+2)-3(2+6)=0`
`rArr lambda^(2)+5lambda - 24=0`
`rArr lambda =- 8 or 3 `