A perpendicular is drawn from a point on the line $\frac{x-1}{2}=\mathrm{ }\frac{y+1}{-1}=\frac{z}{1}$ to the plane $x+y+z=3$ such that the foot of the perpendicular $Q$ also lies on the plane $x-y+z=3$. Then the coordinates of $Q$ are

The given line is $\frac{x-1}{2}=\frac{y+1}{-1}=\frac{z}{1}=\lambda , ($let$)$

$\Rightarrow \frac{x-1}{2}=\lambda , \frac{y+1}{-1}=\lambda , \frac{z}{1}=\lambda$

$\Rightarrow x=2\lambda +1, y=-\lambda -1, z=\lambda .$

Hence, any point on the line is $\left(2\lambda +1, -\lambda -1, \lambda \right).$

Thus, the point $P$ on the line is $\left(2\lambda +1, -\lambda -1, \lambda \right).$

We know that the foot of perpendicular from a point $\left(x_1, y_1, z_1\right)$ on the plane $ax+by+cz+d=0$ is $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}=-\frac{\left(a{x}_1+b{y}_1+c{z}_1\right)}{a^2+b^2+c^2}.$

Thus, the foot of perpendicular $Q$ from $P$ on the plane $x+y+z-3=0$ is given by

$\frac{x-2\lambda -1}{1}=\frac{y+\lambda +1}{1}=\frac{z-\lambda }{1}=-\frac{\left(2\lambda -3\right)}{3}$

$\because Q$ lies on $x+y+z=3 \& x-y+z=3$

On adding the two equations, we get $y=0$

$\Rightarrow \frac{\lambda +1}{1}=\frac{-2\lambda +3}{3}$

$\Rightarrow \lambda =0$

Hence, we have $\frac{x-1}{1}=\frac{z}{1}=-\frac{\left(-3\right)}{3}$

$\Rightarrow x=2, z=1$

So, the coordinate of $Q$ are $\left(2, 0, 1\right).$