Let the vertices Q and R of the triangle PQR | vedclass

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(B) Let $M$ be the foot of the perpendicular from $P(0,2,3)$ to the line $L: \frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3} = \lambda$.Any point on the lin

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$2 m - 5 \sqrt{21} n = 0$

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(B) Let $M$ be the foot of the perpendicular from $P(0,2,3)$ to the line $L: \frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3} = \lambda$.Any point on the line is $M(5\lambda-3, 2\lambda+1, 3\lambda-4)$.The direction ratios of $PM$ are $(5\lambda-3-0, 2\lambda+1-2, 3\lambda-4-3) = (5\lambda-3, 2\lambda-1, 3\lambda-7)$.Since $PM \perp L$,the dot product of the direction ratios of $PM$ and $L$ is zero:$5(5\lambda-3) + 2(2\lambda-1) + 3(3\lambda-7) = 0$$25\lambda - 15 + 4\lambda - 2 + 9\lambda - 21 = 0$$38\lambda - 38 = 0 \Rightarrow \lambda = 1$.Thus,$M = (5(1)-3, 2(1)+1, 3(1)-4) = (2, 3, -1)$.The length of the altitude $PM$ is $\sqrt{(2-0)^2 + (3-2)^2 + (-1-3)^2} = \sqrt{4+1+16} = \sqrt{21}$.The area of $	riangle PQR = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes QR 	imes PM = \frac{1}{2} 	imes 5 	imes \sqrt{21} = \frac{5\sqrt{21}}{2}$.Given area $= \frac{m}{n} = \frac{5\sqrt{21}}{2}$.Comparing,we get $2m = 5\sqrt{21}n$,which implies $2m - 5\sqrt{21}n = 0$.

Let the vertices $Q$ and $R$ of the triangle $PQR$ lie on the line $\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}$. Given $QR=5$ and the coordinates of the point $P$ are $(0,2,3)$. If the area of the triangle $PQR$ is $\frac{m}{n}$,then:

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