If the lines frac{x-1}{2}=frac{y+1}{k}=frac{z | vedclass

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(B) The lines are given by $L_1: \frac{x-1}{2}=\frac{y+1}{k}=\frac{z}{2}$ and $L_2: \frac{x+1}{5}=\frac{y+1}{2}=\frac{z}{k}$.Since the lines are copla

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$x + 4y - z + 3 = 0$

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(B) The lines are given by $L_1: \frac{x-1}{2}=\frac{y+1}{k}=\frac{z}{2}$ and $L_2: \frac{x+1}{5}=\frac{y+1}{2}=\frac{z}{k}$.Since the lines are coplanar,the condition is $\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \ a_1 & b_1 & c_1 \ a_2 & b_2 & c_2 \end{vmatrix} = 0$.Here,$(x_1, y_1, z_1) = (1, -1, 0)$ and $(x_2, y_2, z_2) = (-1, -1, 0)$.So,$\begin{vmatrix} -1-1 & -1-(-1) & 0-0 \ 2 & k & 2 \ 5 & 2 & k \end{vmatrix} = 0 \implies \begin{vmatrix} -2 & 0 & 0 \ 2 & k & 2 \ 5 & 2 & k \end{vmatrix} = 0$.Expanding along the first row: $-2(k^2 - 4) = 0 \implies k^2 = 4 \implies k = \pm 2$.Case $1$: If $k=2$,the lines are $\frac{x-1}{2}=\frac{y+1}{2}=\frac{z}{2}$ and $\frac{x+1}{5}=\frac{y+1}{2}=\frac{z}{2}$. The normal vector is $\vec{n} = \vec{v_1} 	imes \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 2 & 2 \ 5 & 2 & 2 \end{vmatrix} = 0\hat{i} + 6\hat{j} - 6\hat{k}$. The plane equation is $0(x-1) + 6(y+1) - 6(z-0) = 0 \implies y - z + 1 = 0$.Case $2$: If $k=-2$,the lines are $\frac{x-1}{2}=\frac{y+1}{-2}=\frac{z}{2}$ and $\frac{x+1}{5}=\frac{y+1}{2}=\frac{z}{-2}$. The normal vector is $\vec{n} = \vec{v_1} 	imes \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -2 & 2 \ 5 & 2 & -2 \end{vmatrix} = 0\hat{i} + 14\hat{j} + 14\hat{k}$. The plane equation is $0(x-1) + 14(y+1) + 14(z-0) = 0 \implies y + z + 1 = 0$.Combining both,the equation is $y \pm z + 1 = 0$.

If the lines $\frac{x-1}{2}=\frac{y+1}{k}=\frac{z}{2}$ and $\frac{x+1}{5}=\frac{y+1}{2}=\frac{z}{k}$ are coplanar,then the equation of the plane containing these lines is:

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