# Let L be line obtained from the intersection of two planes $x+2y+z=6$ and $y+2z=4$. If point P $(\alpha ,\beta ,\gamma )$ is the force of perpendicular from (3,2,1) on L, then the value of $21(\alpha +\beta +\gamma )$ equals: Option: 1 102 Option: 2 136 Option: 3 68 Option: 4 142

Given the equation of two plane

x + 2y + z = 6

y + 2z = 4

solving above equation, we get

$x=3 z-2 \text{ and } y=4-2 z$

⇒ The line of intersection of two planes is

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

$\\\because \text{AP}\perp \;\text{to line}$

$\\ \therefore \vec{\mathrm{AP}} \cdot(3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}})=0 \\ (3 \lambda-5) \cdot 3+(-2 \lambda+2)(-2)+(\lambda-1) \cdot 1=0 \\ 9 \lambda-15+4 \lambda-4+\lambda-1=0 \\ 14 \lambda=20 \\ \lambda=\frac{10}{7} \Rightarrow \mathrm{P}\left(\frac{16}{7}, \frac{8}{7}, \frac{10}{7}\right)$

$\\\Rightarrow \alpha+\beta+\gamma=\frac{16+8+10}{7}=\frac{34}{7} \\ 21(\alpha+\beta+\gamma)=102$

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