If for some alpha in R,the lines L_1: frac{x+ | vedclass

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(B) The lines $L_1$ and $L_2$ are coplanar if the determinant of the vector connecting a point on each line and their direction vectors is zero. Given

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$(2,-10,-2)$

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(B) The lines $L_1$ and $L_2$ are coplanar if the determinant of the vector connecting a point on each line and their direction vectors is zero. Given points: $P_1 = (-1, 2, 1)$ and $P_2 = (-2, -1, -1)$. Direction vectors: $\vec{v_1} = (2, -1, 1)$ and $\vec{v_2} = (\alpha, 5-\alpha, 1)$. The condition for coplanarity is: $\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \ a_1 & b_1 & c_1 \ a_2 & b_2 & c_2 \end{vmatrix} = 0$ $\begin{vmatrix} -2-(-1) & -1-2 & -1-1 \ 2 & -1 & 1 \ \alpha & 5-\alpha & 1 \end{vmatrix} = 0$ $\begin{vmatrix} -1 & -3 & -2 \ 2 & -1 & 1 \ \alpha & 5-\alpha & 1 \end{vmatrix} = 0$ Expanding the determinant: $-1(-1 - (5-\alpha)) + 3(2 - \alpha) - 2(2(5-\alpha) - (-1)\alpha) = 0$ $-1(-6+\alpha) + 6 - 3\alpha - 2(10 - 2\alpha + \alpha) = 0$ $6 - \alpha + 6 - 3\alpha - 20 + 2\alpha = 0$ $-2\alpha - 8 = 0 \Rightarrow \alpha = -4$. Substituting $\alpha = -4$ into $L_2$: $L_2: \frac{x+2}{-4} = \frac{y+1}{5-(-4)} = \frac{z+1}{1} \Rightarrow \frac{x+2}{-4} = \frac{y+1}{9} = \frac{z+1}{1}$. Checking option $(B) (2, -10, -2)$: $\frac{2+2}{-4} = \frac{-10+1}{9} = \frac{-2+1}{1} \Rightarrow -1 = -1 = -1$. Thus,the line $L_2$ passes through $(2, -10, -2)$.

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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