Answer:
Option D
Explanation:
Concept involved
To find the foot of perpendicular and find its locus
Formula used
Foot of perpendicular form $(x_{1},y_{1},z_{1})$ to ax+by+cz+d=0 be $(x_{2},y_{2},z_{2})$
then
$\frac{x_{2}-x_{1}}{a}=\frac{y_{2}-y_{1}}{b}=\frac{z_{2}-z_{1}}{a}$
= $\frac{-(ax_{1}+by_{1}+cz_{1}+d)}{a^{2}+b^{2}+c^{2}}$
Any point on, $\frac{x+2}{2}=\frac{y+1}{-1}=\frac{z}{3}=\lambda$
$\Rightarrow x=2\lambda-2,y=\lambda-1, z=3\lambda$
Let focus of perpendicular from $(2\lambda-2,\lambda-1, 3\lambda)$
to x+y+z=3 be ($(x_{2},y_{2},z_{2})$)
$\therefore$ $\frac{x_{2}-(2\lambda-2)}{1}=\frac{y_{2}-(-\lambda-1)}{1}=\frac{z_{2}-(3\lambda)}{1}$
= $-\frac{(2\lambda-2-\lambda-1+3\lambda-3)}{1+1+1}$
= $x_{2}-2\lambda+2=y_{2}+\lambda+1$
= $z_{2}-3\lambda=2-\frac{4\lambda}{3}$
$\therefore$ $x_{2}=\frac{2\lambda}{3}$ , $y_{2}=1-\frac{7\lambda}{3}$, $z_{2}=2+\frac{5\lambda}{3}$
$\Rightarrow \lambda=\frac{x_{2}-0}{2/3}=\frac{y_{2}-1}{-7/3}=\frac{z_{2}-2}{5/3}$
$\therefore$ foot of perpendicular lie on
$\frac{x}{2/3}=\frac{y_{}-1}{-7/3}=\frac{z_{}-2}{5/3}$
$\Rightarrow\frac{x}{2}=\frac{y_{}-1}{-7}=\frac{z_{}-2}{5}$
