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(D) The equations of the planes are $x+2y+z=6$ and $y+2z=4$. To find the line of intersection,we express $x$ and $y$ in terms of $z$. From $y+2z=4$,we

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$102$

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(D) The equations of the planes are $x+2y+z=6$ and $y+2z=4$. To find the line of intersection,we express $x$ and $y$ in terms of $z$. From $y+2z=4$,we get $y=4-2z$. Substituting this into the first plane equation: $x+2(4-2z)+z=6 \Rightarrow x+8-4z+z=6 \Rightarrow x-3z=-2 \Rightarrow x=3z-2$. Thus,the line $L$ can be written in symmetric form as $\frac{x+2}{3} = \frac{y-4}{-2} = z = \lambda$. Any point $P$ on the line $L$ is given by $(3\lambda-2, -2\lambda+4, \lambda)$. The direction vector of the line is $\vec{v} = 3\hat{i}-2\hat{j}+\hat{k}$. Let $A = (3,2,1)$. The vector $\vec{AP} = (3\lambda-2-3, -2\lambda+4-2, \lambda-1) = (3\lambda-5, -2\lambda+2, \lambda-1)$. Since $\vec{AP} \perp \vec{v}$,their dot product is zero: $(3\lambda-5)(3) + (-2\lambda+2)(-2) + (\lambda-1)(1) = 0$. $9\lambda - 15 + 4\lambda - 4 + \lambda - 1 = 0 \Rightarrow 14\lambda - 20 = 0 \Rightarrow \lambda = \frac{10}{7}$. Substituting $\lambda$ back,$P = (3(\frac{10}{7})-2, -2(\frac{10}{7})+4, \frac{10}{7}) = (\frac{16}{7}, \frac{8}{7}, \frac{10}{7})$. Thus,$\alpha+\beta+\gamma = \frac{16+8+10}{7} = \frac{34}{7}$. The value of $21(\alpha+\beta+\gamma) = 21 	imes \frac{34}{7} = 3 	imes 34 = 102$.

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

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