Find the foot of perpendicular from P(1, 2, - 3) to the line $\frac{\mathrm{x}+1}{2}=\frac{\mathrm{y}-3}{-2}=\frac{\mathrm{z}}{-1}$ Also, find the image of P in the given line.

$\text{ Any point on the given line is }(2\mathrm{\lambda }-1,-2\mathrm{\lambda }+3,-\mathrm{\lambda })\text{. }$
$\text{ Therefore, coordinates of }\mathrm{Q}\text{ are }(2\mathrm{\lambda }-1,-2\mathrm{\lambda }+3,-\mathrm{\lambda })\text{. }$
$\text{ Now, }\overrightarrow{\mathrm{PQ}}=(2\mathrm{\lambda }-2)\widehat{\mathrm{i}}+(-2\mathrm{\lambda }+1)\widehat{\mathrm{j}}+(-\mathrm{\lambda }+3)\widehat{\mathrm{k}}$

$\mathrm{Since}, \overrightarrow{\mathrm{PQ}} \mathrm{is} \mathrm{perpendicular} \mathrm{to} \mathrm{the} \mathrm{line}$
$\frac{\mathrm{x}+1}{2}=\frac{\mathrm{y}-3}{-2}=\frac{\mathrm{z}}{-1}$
$\therefore 2(2\mathrm{\lambda }-2)-2(-2\mathrm{\lambda }+1)-1(-\mathrm{\lambda }+3)=0\Rightarrow \mathrm{\lambda }=1$
$\therefore \text{ Foot of the perpendicular is }\mathrm{Q}(1,1,-1)\text{. }$
$\text{ Let }{\mathrm{P}}^{\mathrm{\prime }}(\mathrm{x},\mathrm{y},\mathrm{z})\text{ be the image of }\mathrm{P}\text{ in the line, then }$
$\text{ Coordinates of }\mathrm{Q}=\text{ Coordinates of mid-point of }{\mathrm{PP}}^{\mathrm{\prime }}$
$\Rightarrow (1,1,-1)=\left(\frac{\mathrm{x}+1}{2},\frac{\mathrm{y}+2}{2},\frac{\mathrm{z}-3}{2}\right)$

$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{\mathrm{x}+1}{2}=1,\frac{\mathrm{y}+2}{2}=1\text{ and }\frac{\mathrm{z}-3}{2}=-1\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{x}=1,\mathrm{y}=0\text{ and }\mathrm{z}=1\end{array}$
Hence, the image is (1,0,1).