A line from the origin meets the lines x−21=y−1−2=z+11 and x−8/32=y+3−1=z−11 at P and Q respectively. If length PQ=α than the value of α2=

P=(S+2−2S+1 S−1) Q=(2t+83−t−3, t+1)∴ O, P, Q are collinear ⇒2+S2t+83=−2S+1−t−3=S−1t+1Solving S=3, t=13 P(5, −5, 2) Q=(103, −103, 43) PQ2=6

A line from the origin meets the lines x−21=y−1−2=z+11 and x−8/32=y+3−1=z−11 at P and Q respectively. If length PQ=α than the value of α2=

P=(S+2−2S+1 S−1) Q=(2t+83−t−3, t+1)∴ O, P, Q are collinear ⇒2+S2t+83=−2S+1−t−3=S−1t+1Solving S=3, t=13 P(5, −5, 2) Q=(103, −103, 43) PQ2=6

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A line from the origin meets the lines $\frac{x - 2}{1} = \frac{y - 1}{- 2} = \frac{z + 1}{1}$ and $\frac{x - 8 / 3}{2} = \frac{y + 3}{- 1} = \frac{z - 1}{1}$ at P and Q respectively. If length $P Q = α$ than the value of $α^{2} =$

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$\sqrt{3}$

answer is C.

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Detailed Solution

$P = (S + 2 - 2 S + 1 S - 1)$
$Q = (2 t + \frac{8}{3} - t - 3, t + 1)$
$∴$ O, P, Q are collinear
$\Rightarrow \frac{2 + S}{2 t + \frac{8}{3}} = \frac{- 2 S + 1}{- t - 3} = \frac{S - 1}{t + 1}$
Solving $S = 3, t = \frac{1}{3}$
$P (5, - 5, 2) Q = (\frac{10}{3}, \frac{- 10}{3}, \frac{4}{3})$
$P Q^{2} = 6$