Question

If the distance between the plane $x-2y+z=d$ 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 $|d|$ is:

Answer 1

Let the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}$ be denoted by $L_1$ and $L_2$
Equation of plane containing $L_1$ and $L_2$ is given by:
$\left|\begin{array}{ccc}x-1& y-2& z-3\\ 2& 3& 4\\ 3& 4& 5\end{array}\right|=0$
$(x-1)(-1)-(y-2)(-2)+(z-3)(-1)=0$
$\Rightarrow 1-x+2y-4+3-z=0$
$\Rightarrow -x+2y-z=0\cdots \left(i\right)$
$\Rightarrow x-2y+z=0\cdots \left(i\right)$
Given plane is $x-2y+z=d\cdots \left(ii\right)$
$\left(i\right)$ and $\left(ii\right)$ are parallel.
Now, distance between planes is $\left|\frac{d}{\sqrt{1+4+1}}\right|=\sqrt{6}$
$\Rightarrow |d|=6$