The equation of the plane containing the line | vedclass

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(B) The lines pass through points $A(1, -1, 0)$ and $B(-1, -1, 0)$. The direction vectors are $\vec{v_1} = 2\hat{i} + \lambda\hat{j} + 2\hat{k}$ and $

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$y \pm z + 1 = 0$

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(B) The lines pass through points $A(1, -1, 0)$ and $B(-1, -1, 0)$. The direction vectors are $\vec{v_1} = 2\hat{i} + \lambda\hat{j} + 2\hat{k}$ and $\vec{v_2} = 5\hat{i} + 2\hat{j} + \lambda\hat{k}$.Since the lines lie in the same plane, the vector $\vec{AB} = -2\hat{i} + 0\hat{j} + 0\hat{k}$ must be coplanar with $\vec{v_1}$ and $\vec{v_2}$. Thus, the scalar triple product $[\vec{AB}, \vec{v_1}, \vec{v_2}] = 0$.$\begin{vmatrix} -2 & 0 & 0 \ 2 & \lambda & 2 \ 5 & 2 & \lambda \end{vmatrix} = -2(\lambda^2 - 4) = 0 \implies \lambda^2 = 4 \implies \lambda = \pm 2$.The normal to the plane is $\vec{n} = \vec{v_1} 	imes \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & \lambda & 2 \ 5 & 2 & \lambda \end{vmatrix} = (\lambda^2 - 4)\hat{i} - (2\lambda - 10)\hat{j} + (4 - 5\lambda)\hat{k}$.For $\lambda = 2$, $\vec{n} = 0\hat{i} + 6\hat{j} - 6\hat{k} = 6(\hat{j} - \hat{k})$. The plane is $0(x-1) + 1(y+1) - 1(z-0) = 0 \implies y - z + 1 = 0$.For $\lambda = -2$, $\vec{n} = 0\hat{i} + 14\hat{j} + 14\hat{k} = 14(\hat{j} + \hat{k})$. The plane is $0(x-1) + 1(y+1) + 1(z-0) = 0 \implies y + z + 1 = 0$.Combining these, the equation is $y \pm z + 1 = 0$.

The equation of the plane containing the lines $\frac{x - 1}{2} = \frac{y + 1}{\lambda} = \frac{z}{2}$ and $\frac{x + 1}{5} = \frac{y + 1}{2} = \frac{z}{\lambda}$ is

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