A straight line drawn from the point P(1,3,2) | vedclass

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(D) The line passing through $P(1,3,2)$ and parallel to $\frac{x-2}{1}=\frac{y-4}{2}=\frac{z-6}{1}$ is given by $\frac{x-1}{1}=\frac{y-3}{2}=\frac{z-2

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$A, B, C$

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(D) The line passing through $P(1,3,2)$ and parallel to $\frac{x-2}{1}=\frac{y-4}{2}=\frac{z-6}{1}$ is given by $\frac{x-1}{1}=\frac{y-3}{2}=\frac{z-2}{1} = \lambda$.Any point on this line is $(\lambda+1, 2\lambda+3, \lambda+2)$.Since this line intersects the plane $L_1: x-y+3z=6$ at $Q$,we have:$(\lambda+1) - (2\lambda+3) + 3(\lambda+2) = 6$$\lambda+1-2\lambda-3+3\lambda+6 = 6$$2\lambda + 4 = 6 \Rightarrow 2\lambda = 2 \Rightarrow \lambda = 1$.Thus,$Q = (1+1, 2(1)+3, 1+2) = (2,5,3)$.The length $PQ = \sqrt{(2-1)^2 + (5-3)^2 + (3-2)^2} = \sqrt{1^2 + 2^2 + 1^2} = \sqrt{6}$. So,$(A)$ is $TRUE$.The line passing through $Q(2,5,3)$ and perpendicular to $L_1: x-y+3z=6$ has direction ratios $(1, -1, 3)$.The equation of this line is $\frac{x-2}{1} = \frac{y-5}{-1} = \frac{z-3}{3} = t$.Any point on this line is $(t+2, 5-t, 3t+3)$.This line intersects $L_2: 2x-y+z=-4$ at $R$:$2(t+2) - (5-t) + (3t+3) = -4$$2t+4-5+t+3t+3 = -4$$6t + 2 = -4 \Rightarrow 6t = -6 \Rightarrow t = -1$.Thus,$R = (-1+2, 5-(-1), 3(-1)+3) = (1,6,0)$. So,$(B)$ is $TRUE$.The centroid of $	riangle PQR$ is $\left(\frac{1+2+1}{3}, \frac{3+5+6}{3}, \frac{2+3+0}{3}ight) = \left(\frac{4}{3}, \frac{14}{3}, \frac{5}{3}ight)$. So,$(C)$ is $TRUE$.The perimeter is $PQ + QR + RP = \sqrt{6} + \sqrt{(1-2)^2 + (6-5)^2 + (0-3)^2} + \sqrt{(1-1)^2 + (6-3)^2 + (0-2)^2} = \sqrt{6} + \sqrt{1+1+9} + \sqrt{0+9+4} = \sqrt{6} + \sqrt{11} + \sqrt{13}$. So,$(D)$ is $TRUE$.

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