Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

The equation od given line is $\frac{\text{x}+2}{2}=\frac{\text{y}+1}{-1}=\frac{\text{z}}{3}=\lambda$
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 $\left(\alpha \text{, }\beta \text{, }\gamma \right)$
$\Rightarrow \frac{\alpha +2}{1}=\frac{\beta +1}{1}=\frac{\gamma -0}{1}=\mu \left(\text{say}\right)$
Also the point lies on the plane, hence $\alpha +\beta +\gamma =3$
$$\begin{gathered} \Rightarrow \mu -2+\mu -1+\mu =3 \\ \Rightarrow \mu =2 \end{gathered}$$

Hence the coordinates of $M$ are
$\Rightarrow \text{M}\left(\text{0, 1, 2}\right)$
Similarly foot of perpendicular from $B(0, -2, 3)$ on plane is N $\left(\frac{2}{3}\text{,}\frac{-4}{3}\text{,}\frac{11}{3}\right)$
So the required equation of MN is $\frac{\text{x}-0}{\frac{2}{3}}=\frac{\text{y}-1}{\frac{-7}{3}}=\frac{\text{z}-2}{\frac{5}{3}}$

$\Rightarrow \frac{\text{x}-0}{2}=\frac{\text{y}-1}{-7}=\frac{\text{z}-2}{5}$