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(D) Any point on the line $\frac{x+2}{2}=\frac{y+1}{-1}=\frac{z}{3}=\lambda$ is given by $(2\lambda-2, -\lambda-1, 3\lambda)$.Let us choose two points

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$\frac{x}{2}=\frac{y-1}{-7}=\frac{z-2}{5}$

Vedclass · Answer

(D) Any point on the line $\frac{x+2}{2}=\frac{y+1}{-1}=\frac{z}{3}=\lambda$ is given by $(2\lambda-2, -\lambda-1, 3\lambda)$.Let us choose two points on this line:For $\lambda=0$,point $A = (-2, -1, 0)$.For $\lambda=1$,point $B = (0, -2, 3)$.The foot of the perpendicular $(x, y, z)$ from a point $(x_0, y_0, z_0)$ to the plane $ax+by+cz+d=0$ is given by $\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c} = -\frac{ax_0+by_0+cz_0+d}{a^2+b^2+c^2}$.For point $A(-2, -1, 0)$ and plane $x+y+z-3=0$:$\frac{x+2}{1} = \frac{y+1}{1} = \frac{z-0}{1} = -\frac{-2-1+0-3}{1^2+1^2+1^2} = -\frac{-6}{3} = 2$.So,$x = 0, y = 1, z = 2$. Point $M = (0, 1, 2)$.For point $B(0, -2, 3)$ and plane $x+y+z-3=0$:$\frac{x-0}{1} = \frac{y+2}{1} = \frac{z-3}{1} = -\frac{0-2+3-3}{1^2+1^2+1^2} = -\frac{-2}{3} = \frac{2}{3}$.So,$x = \frac{2}{3}, y = \frac{2}{3}-2 = -\frac{4}{3}, z = \frac{2}{3}+3 = \frac{11}{3}$. Point $N = (\frac{2}{3}, -\frac{4}{3}, \frac{11}{3})$.The line passing through $M(0, 1, 2)$ and $N(\frac{2}{3}, -\frac{4}{3}, \frac{11}{3})$ has direction ratios proportional to $(\frac{2}{3}-0, -\frac{4}{3}-1, \frac{11}{3}-2) = (\frac{2}{3}, -\frac{7}{3}, \frac{5}{3})$,which is equivalent to $(2, -7, 5)$.The equation of the line is $\frac{x-0}{2} = \frac{y-1}{-7} = \frac{z-2}{5}$.

Perpendiculars are drawn from points on the line $\frac{x+2}{2}=\frac{y+1}{-1}=\frac{z}{3}$ to the plane $x+y+z=3$. The feet of the perpendiculars lie on the line:

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