If for some α∈R, the lines L1:x+12=y-2-1=z-11 and L2:x+2α=y+15-α=z+11 are coplanar, then the line L2 passes through the point:

A=(-1,2,1),C=(-2,-1,-1)L1,L2 are coplanar ⇒-1+22+11+12-11α5-α1=0⇒α=-4 Equation of L2 is x+2-4=y+19=z+11=tAny point on L2 is −4t−2,9t−1,t−1=2,−10−2,for t=−1

If for some α∈R, the lines L1:x+12=y-2-1=z-11 and L2:x+2α=y+15-α=z+11 are coplanar, then the line L2 passes through the point:

A=(-1,2,1),C=(-2,-1,-1)L1,L2 are coplanar ⇒-1+22+11+12-11α5-α1=0⇒α=-4 Equation of L2 is x+2-4=y+19=z+11=tAny point on L2 is −4t−2,9t−1,t−1=2,−10−2,for t=−1

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Offline Centres

$If for some α \in R, the lines L_{1} : \frac{x + 1}{2} = \frac{y - 2}{- 1} = \frac{z - 1}{1} and L_{2} : \frac{x + 2}{α} = \frac{y + 1}{5 - α} = \frac{z + 1}{1} are coplanar,$ $then the line L_{2} passes through the point:$

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$(10, - 2, - 2)$

$(- 2, 10, 2)$

$(10, 2, 2)$

$(2, - 10, - 2)$

answer is C.

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Detailed Solution

$\begin{array}{l} A = (- 1, 2, 1), C = (- 2, - 1, - 1) \\ L_{1}, L_{2} are coplanar \end{array}$

$\begin{array}{l} \Rightarrow |\begin{matrix} - 1 + 2 & 2 + 1 & 1 + 1 \\ 2 & - 1 & 1 \\ α & 5 - α & 1 \end{matrix}| = 0 \Rightarrow α = - 4 \\ Equation of L_{2} is \\ \frac{x + 2}{- 4} = \frac{y + 1}{9} = \frac{z + 1}{1} = t \end{array}$