The value of lambda for which the straight li | vedclass

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(D) For a line $\frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}}$ to lie on the plane $Ax+By+Cz=D$, two conditions must be satisfied:

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there is no such $\lambda$

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(D) For a line $\frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}}$ to lie on the plane $Ax+By+Cz=D$, two conditions must be satisfied:
$(i)$ The line must be perpendicular to the normal of the plane: $a_{1}A+b_{1}B+c_{1}C=0$.
(ii) Any point on the line must satisfy the plane equation: $Ax_{1}+By_{1}+Cz_{1}=D$.
Given the line $\frac{x-\lambda}{3}=\frac{y-1}{2+\lambda}=\frac{z-3}{-1}$ and the plane $x-2y+0z=0$:
Condition $(i)$: $3(1) + (2+\lambda)(-2) + (-1)(0) = 0$
$3 - 4 - 2\lambda = 0 \Rightarrow -1 - 2\lambda = 0 \Rightarrow \lambda = -\frac{1}{2}$.
Condition (ii): The point $(\lambda, 1, 3)$ must lie on the plane $x-2y=0$.
Substituting the point: $\lambda - 2(1) = 0 \Rightarrow \lambda = 2$.
Since the two conditions yield different values for $\lambda$ ($\lambda = -\frac{1}{2}$ and $\lambda = 2$), there is no such $\lambda$ for which the line lies on the plane.

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 $x-2y=0$ is

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