Perpendiculars are drawn from points on the line x + 2 2 = y + 1 -...

The equation od given line is $\dfrac{\text{x} + 2}{2} = \dfrac{\text{y} + 1}{- 1} = \dfrac{\text{z}}{3} = \lambda$
Now any two points on this line are
$\text{A}$\left(- 2\text{,} - 1\text{,}0\right)\text{,B}$\left(0\text{,} - 2\text{,}3\right)$ for$\lambda = 0\text{,}1$
Let foot of perpendicular from $A ( - 2, - 1, 0 )$ on plane is \left(\alpha \text{,} \beta \text{,} \gamma\right)$<br />$⇒ \dfrac{\alpha + 2}{1} = \dfrac{\beta + 1}{1} = \dfrac{\gamma - 0}{1} = \mu\left(\text{say}\right)$<br />Also the point lies on the plane, hence$\alpha + \beta + \gamma = 3$<br />$⇒ \mu - 2 + \mu - 1 + \mu = 3 ⇒ \mu = 2$</p><p>Hence the coordinates of$M$are<br />$⇒ \text{M}\left(\text{0, 1, 2}\right)$<br />Similarly foot of perpendicular from$B ( 0, - 2, 3 )$on plane is N \left(\dfrac{2}{3} \text{,} \dfrac{- 4}{3} \text{,} \dfrac{1 1}{3}\right)$
So the required equation of MN is $\dfrac{\text{x} - 0}{\dfrac{2}{3}} = \dfrac{\text{y} - 1}{\dfrac{- 7}{3}} = \dfrac{\text{z} - 2}{\dfrac{5}{3}}$

$⇒ \dfrac{\text{x} - 0}{2} = \dfrac{\text{y} - 1}{- 7} = \dfrac{\text{z} - 2}{5}$