The foot of the perpendicular drawn from the point (4,2,3) to the line joining the points

Equation of $\mathrm{AB}=\overrightarrow{\mathrm{r}}=(\hat{\mathrm{i}}+\hat{\mathrm{j}})+\lambda(3 \hat{\mathrm{j}}-3 \hat{\mathrm{k}})$

Let coordinates of M $=(1,(1+3 \lambda),-3 \lambda)$.

$\overrightarrow{\mathrm{PM}}=-3 \hat{\mathrm{i}}+(3 \lambda-1) \hat{\mathrm{j}}-3(\lambda+1) \hat{\mathrm{k}}$

$\overline{\mathrm{AB}}=3 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}$

$\Rightarrow \lambda=-\frac{1}{3}$

$\therefore \quad M=(1,0,1)$

Clearly $\mathrm{M}$ lies on $2 \mathrm{x}+\mathrm{y}-\mathrm{z}=1$.