Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Equation of the line through $(1,-2,3)$ parallel to the line $\frac{x}{2}=\frac{y}{3}=\frac{z-1}{-6}$ is $\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{-6}=r$ (say) Then any point on (1) is $(2r+1,3r-2,-6r$ $+3$ ) If this point lies on the plane $x-y+z=5$ then $$(2r+1)-(3r-2)+(-6r+3)=5\Rightarrow r=\frac{1}{7}$$ Hence the point is $\left(\frac{9}{7},-\frac{11}{7},\frac{15}{7}\right)$ Distance between $(1,-2,3)$ and $$\begin{array}{l}\left(\frac{9}{7},-\frac{11}{7},\frac{15}{7}\right)\\ =\sqrt{\left(\frac{4}{49}+\frac{9}{49}+\frac{36}{49}\right)}=\sqrt{\left(\frac{49}{49}\right)}=1\end{array}$$