Question

If the line $\frac{x-{\lambda }^2}{-3\lambda }=\frac{y+5}{\lambda }=\frac{z+3}{-1}$ lies on the plane $x+\lambda y-2z=0,$ then the value of $\lambda$ is equal to

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B $2$

Given :
Line $:\frac{x-{\lambda }^2}{-3\lambda }=\frac{y+5}{\lambda }=\frac{z+3}{-1}$
passes through point $P({\lambda }^2,-5,-3)$
And $D.r^{\prime }s$ of line $(a,b,c)=(-3\lambda ,\lambda ,-1)$
So, point $P$ satisfy the plane $x+\lambda y-2z=0,$
$\Rightarrow {\lambda }^2+\lambda (-5)-2(-3)=0\Rightarrow (\lambda -2)(\lambda -3)=0\Rightarrow \lambda =2,3\cdots \left(i\right)$
Also,
$\Rightarrow D.r^{\prime }s$ of normal to the plane is ; $(a_1,b_1,c_1)=(1,\lambda ,-2)$
$\Rightarrow a{a}_1+b{b}_1+c{c}_1=0$
$\Rightarrow -3\lambda +\lambda \left(\lambda \right)-2(-1)=0\Rightarrow (\lambda -1)(\lambda -2)=0\Rightarrow \lambda =1,2\cdots \left(ii\right)$
Taknig the common value of $\left(i\right)$ and $\left(ii\right):\lambda =2$