# Let L be a

Question:

Let $L$ be a line obtained from the intersection of two planes $x+2 y+z=6$ and $y+2 z=4$. If point $\mathrm{P}(\alpha, \beta, \gamma)$ is the foot of perpendicular from $(3,2,1)$ on $\mathrm{L}$, then the value of $21(\alpha+\beta+\gamma)$ equals :

1. 142

2. 68

3. 136

4. 102

Correct Option: , 4

Solution:

$x+2 y+z=6$

(y+2 z=4) \times 2

$x-3 z=-2 \Rightarrow x=3 z-2 \Rightarrow y=4-2 z$

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

$\Rightarrow$ line of intersection of two planes is

$\frac{x+2}{3}=\frac{y-4}{-2}=z=\lambda$

$\because \mathrm{AP} \perp^{\text {ar }}$ to line

$\therefore \overline{\mathrm{AP}} \cdot(3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}})=0$

$(3 \lambda-5) \cdot 3+(-2 \lambda+2)(-2)+(\lambda-1) \cdot 1=0$

$9 \lambda-15+4 \lambda-4+\lambda-1=0$

$14 \lambda=20$

$\lambda=\frac{10}{7} \Rightarrow \mathrm{P}\left(\frac{16}{7}, \frac{8}{7}, \frac{10}{7}\right)$

$\Rightarrow \alpha+\beta+\gamma=\frac{16+8+10}{7}=\frac{34}{7}$

$21(\alpha+\beta+\gamma)=102$