1. Key Concepts and Formulas

• Equation of a Straight Line in 3D: A straight line passing through a point $\vec{a} = (x_0, y_0, z_0)$ and having a direction vector $\vec{b} = (l, m, n)$ can be represented in vector form as $\vec{r} = \vec{a} + k\vec{b}$, or in Cartesian form as $\frac{x-x_0}{l} = \frac{y-y_0}{m} = \frac{z-z_0}{n} = k$.
• Condition for Coplanarity of Two Lines: Two straight lines, given in vector form as $\vec{r_1} = \vec{a_1} + s\vec{b_1}$ and $\vec{r_2} = \vec{a_2} + t\vec{b_2}$, are coplanar if and only if the scalar triple product of the vectors $(\vec{a_2} - \vec{a_1})$, $\vec{b_1}$, and $\vec{b_2}$ is zero. This means $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = 0$.
• Determinant Form for Coplanarity: In Cartesian coordinates, if the first line passes through $(x_1, y_1, z_1)$ with direction ratios $(l_1, m_1, n_1)$ and the second line passes through $(x_2, y_2, z_2)$ with direction ratios $(l_2, m_2, n_2)$, they are coplanar if: $\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \end{vmatrix} = 0$

2. Step-by-Step Solution

Step 1: Extract points and direction vectors for Line 1. The first line is given by $x=1+s, y=-3-\lambda s, z=1+\lambda s$. To find a point on the line, we can set $s=0$. This gives the point $\vec{a_1} = (1, -3, 1)$. The direction vector of the line is given by the coefficients of $s$. Based on the problem structure, we interpret the direction vector as $\vec{b_1} = (\lambda, -\lambda, \lambda)$. This is a common pattern in such problems where the parameter $\lambda$ influences the overall direction. So, $(x_1, y_1, z_1) = (1, -3, 1)$ and $(l_1, m_1, n_1) = (\lambda, -\lambda, \lambda)$.

Step 2: Extract points and direction vectors for Line 2. The second line is given by $x = {t \over 2}, y = 1 + t, z = 2 - t$. To find a point on the line, we can set $t=0$. This gives the point $\vec{a_2} = (0, 1, 2)$. The direction vector of the line is given by the coefficients of $t$. So, $(x_2, y_2, z_2) = (0, 1, 2)$ and $(l_2, m_2, n_2) = (1/2, 1, -1)$.

Step 3: Calculate the vector connecting the two points. We need the vector $(\vec{a_2} - \vec{a_1})$, which corresponds to $(x_2-x_1, y_2-y_1, z_2-z_1)$. $(x_2-x_1, y_2-y_1, z_2-z_1) = (0-1, 1-(-3), 2-1) = (-1, 4, 1)$

Step 4: Apply the coplanarity condition. For the two lines to be coplanar, the determinant formed by the vector $(\vec{a_2} - \vec{a_1})$ and the direction vectors $\vec{b_1}$ and $\vec{b_2}$ must be zero. $\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \end{vmatrix} = 0$ Substitute the values obtained in Step 1, 2, and 3: $\begin{vmatrix} -1 & 4 & 1 \\ \lambda & -\lambda & \lambda \\ 1/2 & 1 & -1 \end{vmatrix} = 0$

Step 5: Evaluate the determinant and solve for $\lambda$. Expand the determinant along the first row: $-1((-\lambda)(-1) - (\lambda)(1)) - 4((\lambda)(-1) - (\lambda)(1/2)) + 1((\lambda)(1) - (-\lambda)(1/2)) = 0$ $-1(\lambda - \lambda) - 4(-\lambda - \lambda/2) + 1(\lambda + \lambda/2) = 0$ $-1(0) - 4(-3\lambda/2) + 1(3\lambda/2) = 0$ $0 + 6\lambda + \frac{3\lambda}{2} = 0$ Combine the terms with $\lambda$: $\frac{12\lambda + 3\lambda}{2} = 0$ $\frac{15\lambda}{2} = 0$ $15\lambda = 0$ $\lambda = 0$

3. Common Mistakes & Tips

• Incorrectly identifying points and direction vectors: Ensure you correctly extract $(x_0, y_0, z_0)$ by setting the parameter to zero, and $(l, m, n)$ as the coefficients of the parameter. Pay close attention to signs.
• Errors in determinant expansion: Be meticulous with signs when expanding the $3 \times 3$ determinant. A single sign error can lead to an incorrect value of $\lambda$.
• Misinterpreting the direction vector: In problems involving parameters like $\lambda$ within the line equations, sometimes a coefficient of $s$ (or $t$) might involve $\lambda$ itself, as interpreted here for the $x$-component of the first line's direction vector. Always double-check how parameters affect direction ratios.

4. Summary

To determine the value of $\lambda$ for which the given lines are coplanar, we first identified a point and the direction vector for each line. Then, we applied the condition for coplanarity of two lines, which states that the scalar triple product of the vector connecting a point on the first line to a point on the second line, and the two direction vectors, must be zero. This condition was expressed as a $3 \times 3$ determinant. Evaluating this determinant and solving the resulting linear equation for $\lambda$ yielded the value $\lambda=0$.

5. Final Answer The final answer is \boxed{0} which corresponds to option (A).