Question

Equation of the plane containing the line $\frac{x-2}{\frac{1}{\sqrt{3}}}=\frac{y-1}{-\frac{1}{\sqrt{3}}}=\frac{z-3}{\frac{1}{\sqrt{3}}}$ is $2(x+1)+\lambda (y-1)+3(z-3)=0$, then value of $\lambda$ equals

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

The correct option is D $5$

Equation of line $\frac{x-2}{\frac{1}{\sqrt{3}}}=\frac{y-1}{\frac{-1}{\sqrt{3}}}=\frac{z-3}{\frac{1}{\sqrt{3}}}$ contains the plan $2(x+1)+\lambda (y-1)+3(z-3)$ hence they are per their direction ratios multiplication sum will be zero so here, $\frac{2}{\sqrt{3}}-\frac{\lambda }{\sqrt{3}}+\frac{3}{\sqrt{3}}=0$

$\Rightarrow \frac{5}{\sqrt{3}}=\frac{\lambda }{\sqrt{3}}$

$\Rightarrow \lambda =5$

So value of $\lambda$ will be $5$