Let the polished side of the mirror be along the line x/1=(1-y)/-2=(2 z-4)/6. A point P (1,6,3), some distance away from the mirror, has its image…

Equation of given line is $\displaystyle \frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$ Let coordinates of point on the line be ( $\displaystyle \lambda, 2 \lambda+1,3 \lambda+2$ ) for some $\displaystyle \lambda$ Drs of line perpendicular to line along mirror are $\displaystyle <\lambda-1,2 \lambda-5,3 \lambda-1> (\lambda-1) \cdot 1+(2 \lambda-5) \cdot 2+(3 \lambda-1) \cdot 3=0$ gives $\displaystyle \lambda=1$ Coordinates of foot of perpendicular are ( $\displaystyle 1,3,5$ ) For image \[\frac{x+1}{2}=1, \frac{y+6}{2}=3, \frac{z+3}{2}=5 \text { gives image as }(1,0,7) \] Required distance $\displaystyle =\sqrt{0+36+16}=2 \sqrt{13}$