If for some alpha in R,the lines L_{1}: frac{ | vedclass

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(D) The lines $L_{1}$ and $L_{2}$ are coplanar if the determinant of the matrix formed by the vector connecting a point on each line and the direction

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$(2, -10, -2)$

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(D) The lines $L_{1}$ and $L_{2}$ are coplanar if the determinant of the matrix formed by the vector connecting a point on each line and the direction vectors of the lines is zero.Point on $L_{1}$ is $P_{1}(-1, 2, 1)$ and direction vector is $\vec{v}_{1} = (2, -1, 1)$.Point on $L_{2}$ is $P_{2}(-2, -1, -1)$ and direction vector is $\vec{v}_{2} = (\alpha, 5-\alpha, 1)$.The vector connecting $P_{1}$ and $P_{2}$ is $\vec{P_{1}P_{2}} = (-2 - (-1), -1 - 2, -1 - 1) = (-1, -3, -2)$.For coplanarity,$\left|\begin{array}{ccc} -1 & -3 & -2 \ 2 & -1 & 1 \ \alpha & 5-\alpha & 1 \end{array}ight| = 0$.Expanding the determinant:$-1(-1 - (5-\alpha)) + 3(2 - \alpha) - 2(2(5-\alpha) - \alpha(-1)) = 0$$-1(\alpha - 6) + 6 - 3\alpha - 2(10 - 2\alpha + \alpha) = 0$$-\alpha + 6 + 6 - 3\alpha - 20 + 2\alpha = 0$$-2\alpha - 8 = 0 \Rightarrow \alpha = -4$.Substituting $\alpha = -4$ into $L_{2}$,we get $\frac{x+2}{-4} = \frac{y+1}{5-(-4)} = \frac{z+1}{1} \Rightarrow \frac{x+2}{-4} = \frac{y+1}{9} = \frac{z+1}{1}$.Checking the options,for point $(2, -10, -2)$:$\frac{2+2}{-4} = \frac{4}{-4} = -1$,$\frac{-10+1}{9} = \frac{-9}{9} = -1$,$\frac{-2+1}{1} = -1$.Since all ratios are equal,the line $L_{2}$ passes through $(2, -10, -2)$.

Column-$I$	Column-$II$
$(a)$ Gymnosperms	$(p)$ Trimerous flowers
$(b)$ Monocots	$(q)$ Absence of flowers
$(c)$ Dicots	$(r)$ Primitive flowers
$(d)$ Pteridophytes	$(s)$ Tetramerous or pentamerous flowers

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

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