Question

If the distance between the plane $Ax-2y+z=d$ and the plane containing the lines

$\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}$ is $\sqrt{6}$, then the value of $|d|$ is

• A. $3$
• B. $4$
• C. $5$
• D. $6$

Answer 1

The correct option is D $6$

From the given equations, we get

$2l+3m+4n=0$

$3l+4m+5n=0$

$\frac{l}{-1}=\frac{m}{2}=\frac{n}{-1}$

Equation of plane will be:

$\Rightarrow a(x-1)+b(y-2)+c(z-3)=0$

$\Rightarrow -1(x-1)+2(y-2)-1(z-3)=0$

$\Rightarrow -x+1+2y-4-z+3=0$

$\Rightarrow -x+2y-z=0$

$\Rightarrow x-2y+z=0$

$\Rightarrow \frac{|d|}{\sqrt{6}}=\sqrt{6}$

$\Rightarrow d=6$