Perpendiculars are drawn from points on the line x+22=y+1−1=z3 to the plane x+y+z=3. The feet of perpendiculars lie on the line

# 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 perpendiculars lie on the line

1. A

$\frac{x}{5}=\frac{y-1}{8}=\frac{z-2}{-13}$

2. B

$\frac{x}{2}=\frac{y-1}{3}=\frac{z-2}{-5}$

3. C

$\frac{x}{4}=\frac{y-1}{3}=\frac{z-2}{-7}$

4. D

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

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### Solution:

Let $\frac{x+2}{2}=\frac{y+1}{-1}=\frac{z}{3}=k$ then is a
point on the line. Let the foot of the perpendicular from this point on the plane  be

$\left(\alpha ,\beta ,\gamma \right)$ then it is given by

$\frac{\alpha -\left(2k-2\right)}{1}=\frac{\beta -\left(k-1\right)}{1}=\frac{\gamma -3k}{1}$

$=-\frac{\left(2k-2\right)+\left(-k-1\right)+\left(3k\right)-3}{3}=\frac{6-4k}{3}$

Thus, $\alpha =\frac{2k}{3},\beta =1-\frac{7k}{3},\gamma =2+\frac{5k}{3}$ then the line is
given by $\frac{x}{2/3}=\frac{y-1}{-7/3}=\frac{z-2}{5/3}$

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