Equation of Plane Containing Coplanar Straight Lines

To determine the plane(s) containing the given straight lines, we first express the equations of the lines in vector form.

The given lines are:

$$L_1: \frac{x-1}{2} = \frac{y+1}{k} = \frac{z}{2}$$ $$L_2: \frac{x+1}{5} = \frac{y+1}{2} = \frac{z}{k}$$

From line $L_1$, a point on the line is $A(1, -1, 0)$ and its direction vector is:

$$\vec{d}_1 = 2\hat{i} + k\hat{j} + 2\hat{k}$$

From line $L_2$, a point on the line is $B(-1, -1, 0)$ and its direction vector is:

$$\vec{d}_2 = 5\hat{i} + 2\hat{j} + k\hat{k}$$

The vector joining points $A$ and $B$ is:

$$\vec{AB} = (-1 - 1)\hat{i} + (-1 - (-1))\hat{j} + (0 - 0)\hat{k} = -2\hat{i} + 0\hat{j} + 0\hat{k}$$

Condition for Coplanarity

Two lines are coplanar if and only if the scalar triple product $[\vec{AB}, \vec{d}_1, \vec{d}_2] = 0$:

$$\begin{vmatrix} -2 & 0 & 0 \\ 2 & k & 2 \\ 5 & 2 & k \end{vmatrix} = 0$$

Expanding the determinant along the first row:

$$-2(k^2 - 4) = 0 \implies k^2 - 4 = 0 \implies k = \pm 2$$

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Case 1: When $k = 2$

The direction vectors are:

$$\vec{d}_1 = \langle 2, 2, 2 \rangle \quad \text{and} \quad \vec{d}_2 = \langle 5, 2, 2 \rangle$$

The normal vector $\vec{n}_1$ to the plane containing both lines is given by the cross product $\vec{d}_1 \times \vec{d}_2$:

$$\vec{n}_1 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 2 & 2 \\ 5 & 2 & 2 \end{vmatrix} = \hat{i}(4 - 4) - \hat{j}(4 - 10) + \hat{k}(4 - 10) = 0\hat{i} + 6\hat{j} - 6\hat{k}$$

Dividing by $6$, we can take the normal vector as:

$$\vec{n}_1 = \langle 0, 1, -1 \rangle$$

The equation of the plane passing through $A(1, -1, 0)$ is:

$$0(x - 1) + 1(y - (-1)) - 1(z - 0) = 0$$ $$(y + 1) - z = 0 \implies y - z = -1$$

This matches Option (C).

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Case 2: When $k = -2$

The direction vectors are:

$$\vec{d}_1 = \langle 2, -2, 2 \rangle \quad \text{and} \quad \vec{d}_2 = \langle 5, 2, -2 \rangle$$

The normal vector $\vec{n}_2$ to the plane containing both lines is:

$$\vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -2 & 2 \\ 5 & 2 & -2 \end{vmatrix} = \hat{i}(4 - 4) - \hat{j}(-4 - 10) + \hat{k}(4 - (-10)) = 0\hat{i} + 14\hat{j} + 14\hat{k}$$

Dividing by $14$, we can take the normal vector as:

$$\vec{n}_2 = \langle 0, 1, 1 \rangle$$

The equation of the plane passing through $A(1, -1, 0)$ is:

$$0(x - 1) + 1(y - (-1)) + 1(z - 0) = 0$$ $$(y + 1) + z = 0 \implies y + z = -1$$

This matches Option (B).

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Conclusion

The planes containing these two lines are:

• $y + z = -1$
• $y - z = -1$

Therefore, the correct options are (B) and (C).