Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

Line passing through $(1,4,0)$ and parallel to $\frac{x}{1}=\frac{y-2}{2}=\frac{z+3}{3}$ is $L:\frac{x-1}{1}=\frac{y-4}{2}=\frac{z}{3}$
Any point on $L:(\lambda +1,2\lambda +4,3\lambda )$
Any point on $\frac{x-2}{2}=\frac{y-6}{3}=\frac{z-3}{4}$ is $(2\mu +2,3\mu +6,4\mu +3)$
$\begin{array}{l}\lambda +1=2\mu +2\\ 2\lambda +4=3\mu +6\\ 3\lambda =4\mu +3\end{array}\}\lambda =1\mu =0$
Point: $(2,6,3)$
$\begin{array}{rl}& \text{ Distance }=\sqrt{(2-1{)}^2+(6-4{)}^2+(3-0{)}^2}\\ & =\sqrt{1+4+9}=\sqrt{14}\end{array}$