Let the line passing through the points (–1, 2, 1) and parallel to the line $\frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}+1}{3}=\frac{\mathrm{z}}{4}$intersect the line $\frac{\mathrm{x}+2}{3}=\frac{\mathrm{y}-3}{2}=\frac{\mathrm{z}-4}{1}$at the point P. Then the distance of P from the point Q(4, – 5, 1) is :

Equation of line through point (–1, 2,1) is →

$\Rightarrow \frac{\mathrm{x}+1}{2}=\frac{\mathrm{y}-2}{3}=\frac{\mathrm{z}-1}{4}-\left(2\right)=\mathrm{\lambda }$ 
So, $\left[\begin{array}{l}\mathrm{x}=2\mathrm{\lambda }-1\\ \mathrm{y}=3\mathrm{\lambda }+2\\ \mathrm{z}=4\mathrm{\lambda }+1\end{array}\right.$
By $\text{ (1) }\to \frac{\mathrm{x}+2}{3}=\frac{\mathrm{y}-3}{2}=\frac{\mathrm{z}-4}{1}=\mathrm{\mu }\text{ (Let) }$
So, $\left[\begin{array}{l}\mathrm{x}=3\mathrm{\mu }-2\\ \mathrm{y}=2\mathrm{\mu }+3\\ \mathrm{z}=\mathrm{\mu }+4\end{array}\right.$
For intersection point ‘P’
$\begin{array}{r}\mathrm{x}=2\mathrm{\lambda }-1=3\mathrm{\mu }-2\\ \mathrm{y}=3\mathrm{\lambda }+2=2\mathrm{\mu }+3\\ \mathrm{z}=4\mathrm{\lambda }+1=\mathrm{\mu }+4\end{array} \left[\stackrel{\mathrm{\lambda }=1}{\mathrm{\mu }=1}\right]$
So, point P(x, y, z) = (1, 5, 5) & Q(4, – 5, 1)
$\begin{array}{l}\therefore \mathrm{PQ}=\sqrt{9+100+16}\\ =\sqrt{125}=5\sqrt{5}\end{array}$