MCQ
The value of $\lambda$ for which the straight line $\frac{x-\lambda}{3}=\frac{y-1}{2+\lambda}=\frac{z-3}{-1}$ may lie on the plane $x-2 y=0$ is
  • A
    1
  • B
    $0$
  • $-\frac{1}{2}$
  • D
    there is no such $\lambda$

Answer

Correct option: C.
$-\frac{1}{2}$
(C)
Let $a , b , c =3,2+\lambda,-1$ and $a _1, b_1, c _1=1,-2,0$
Since the line lies on the plane,
$aa _1+ bb _1+ cc _1=0$
$\Rightarrow 3(1)+(2+\lambda)(-2)+(-1)(0)=0$
$\Rightarrow \lambda=\frac{-1}{2}$

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