Question

The distance of the point $(1,1,1)$ from the plane passing through the point $(-1,-2,-1)$ and whose normal is perpendicular to both the lines $L_1$ and $L_2$ is

• A. $\frac{2}{\sqrt{75}}$
• B. $\frac{7}{\sqrt{75}}$
• C. $\frac{13}{\sqrt{75}}$
• D. $\frac{23}{\sqrt{75}}$

Answer 1

The correct option is C $\frac{13}{\sqrt{75}}$

$L_1=\frac{x+1}{3}=\frac{y+2}{1}=\frac{z+1}{2}.......\left(i\right)$

$L_2=\frac{x-2}{1}=\frac{y+2}{2}=\frac{z-3}{3}.........\left(ii\right)$

Equation of plane passing through $(-1,-2,-1)$

$a(x+1)+b(y+2)+c(z+1)=0........\left(iii\right)$

Normal of $\left(iii\right)$ is perpendicular to $\left(i\right)$ and $\left(ii\right)$

$3a+b+2c=0...........\left(iv\right)$

$a+2b+3c=0..............\left(v\right)$

Solving $\left(iv\right)$ and $\left(v\right)$

$\frac{a}{3-4}=\frac{b}{2-9}=\frac{c}{6-1}=k$

$\Rightarrow \frac{a}{-1}=\frac{b}{-7}=\frac{c}{5}=k$

$\Rightarrow a=-k$

$b=-7k$

$c=5k$

Putting $a,b,c$ in $\left(iii\right)$

$-k(x+1)-7k(y+2)+5k(z+1)=0$

$\Rightarrow -k[(x+1)+7(y+2)-5(z+1)]=0$

$\Rightarrow (x+1)+7(y+2)-5(z+1)=0$

$\Rightarrow x+1+7y+14-5z-5=0$

$\Rightarrow x+7y-5z+10=0$

Required distance

$=\frac{1+7-5+10}{\sqrt{1^2+7^2+5^2}}=\frac{13}{\sqrt{75}}$