If the distance between the plane $A x-2 y+z=d$ and the plane containing the lines…
If the distance between the plane $A x-2 y+z=d$ and the plane containing the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4} \quad$ and $\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}$ is $\sqrt{6}$, then $|d|$ is
Solution
Equation of plane containing the given lines is $\left|\begin{array}{ccc}x-1 & y-2 & z-3 \\ 2 & 3 & 4 \\ 3 & 4 & 5\end{array}\right|=0$
$
\begin{array}{rr}
\Rightarrow & (x-1)(-1)-(y-2)(-2) \\
& +(z-3)(-1)=0 \\
\Rightarrow & -x+1+2 y-4-z+3=0 \\
\Rightarrow & -x+2 y-z=0 \quad \ldots(i)
\end{array}
$
Given plane is
$
x-2 y+z=d
$
Eqs. (i) and (ii) are parallel.
Clearly, $\quad A=1$
Now, distance between plane
$
\begin{aligned}
& =\left|\frac{d}{\sqrt{1+4+1}}\right|=\sqrt{6} \\
& \Rightarrow \quad|d|=6 \\
&
\end{aligned}
$