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(D) Let $M$ be the foot of the perpendicular from $P(1, 2, 3)$ to the line $L$. Any point on the line $L$ is given by $M(3\lambda+6, 2\lambda+1, 3\lam

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$125$

Vedclass · Answer

(D) Let $M$ be the foot of the perpendicular from $P(1, 2, 3)$ to the line $L$. Any point on the line $L$ is given by $M(3\lambda+6, 2\lambda+1, 3\lambda+2)$.The direction vector of the line $L$ is $\vec{v} = 3\hat{i} + 2\hat{j} + 3\hat{k}$.The vector $\vec{PM} = (3\lambda+6-1)\hat{i} + (2\lambda+1-2)\hat{j} + (3\lambda+2-3)\hat{k} = (3\lambda+5)\hat{i} + (2\lambda-1)\hat{j} + (3\lambda-1)\hat{k}$.Since $\vec{PM} \perp \vec{v}$,their dot product is zero:$3(3\lambda+5) + 2(2\lambda-1) + 3(3\lambda-1) = 0$$9\lambda + 15 + 4\lambda - 2 + 9\lambda - 3 = 0$$22\lambda + 10 = 0 \implies \lambda = -\frac{5}{11}$.Substituting $\lambda$ into $M$,we get $M = (3(-\frac{5}{11})+6, 2(-\frac{5}{11})+1, 3(-\frac{5}{11})+2) = (\frac{51}{11}, \frac{1}{11}, \frac{7}{11})$.Since $Q$ is the image of $P$ in the line,$M$ is the midpoint of $PQ$. Thus,$M = \frac{P+Q}{2} \implies Q = 2M - P = (2(\frac{51}{11})-1, 2(\frac{1}{11})-2, 2(\frac{7}{11})-3) = (\frac{91}{11}, -\frac{20}{11}, -\frac{19}{11})$.$R$ divides $PQ$ in the ratio $1:3$,so $R = \frac{1(Q) + 3(P)}{1+3} = \frac{Q + 3P}{4}$.$R = \frac{1}{4} ((\frac{91}{11} + 3), (-\frac{20}{11} + 6), (-\frac{19}{11} + 9)) = \frac{1}{4} (\frac{124}{11}, \frac{46}{11}, \frac{80}{11}) = (\frac{31}{11}, \frac{23}{22}, \frac{20}{11})$.$\alpha+\beta+\gamma = \frac{62+23+40}{22} = \frac{125}{22}$.$22(\alpha+\beta+\gamma) = 125$.

Let the image of the point $P(1, 2, 3)$ in the line $L : \frac{x-6}{3} = \frac{y-1}{2} = \frac{z-2}{3}$ be $Q$. Let $R(\alpha, \beta, \gamma)$ be a point that divides the line segment $PQ$ internally in the ratio $1:3$. Then the value of $22(\alpha+\beta+\gamma)$ is equal to

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