1. Key Concepts and Formulas

   • Coplanarity of Two Lines: Two lines in 3D space are coplanar if they lie in the same plane. This occurs if they are parallel and distinct, identical, or intersecting.
   • Symmetric Form of a Line: A line passing through a point $P_1(x_1, y_1, z_1)$ with direction vector $\vec{d_1}=(a_1, b_1, c_1)$ is given by $\frac{x-x_1}{a_1} = \frac{y-y_1}{b_1} = \frac{z-z_1}{c_1}$
   • Condition for Coplanarity (Determinant Form): Given two lines $L_1$ and $L_2$ passing through points $P_1(x_1, y_1, z_1)$ and $P_2(x_2, y_2, z_2)$ respectively, with direction vectors $\vec{d_1}=(a_1, b_1, c_1)$ and $\vec{d_2}=(a_2, b_2, c_2)$, they are coplanar if and only if the scalar triple product of the vector $\vec{P_1P_2}$ and their direction vectors $\vec{d_1}$, $\vec{d_2}$ is zero. This is expressed as: $\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = 0$ This condition covers all cases of coplanarity, including intersecting lines and parallel lines (distinct or identical).

2. Step-by-Step Solution

   Step 1: Identify Points and Direction Vectors from the Given Lines We extract the coordinates of a point and the direction ratios for each line from their symmetric forms.

   For Line 1 ($L_1$): $\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z + 3}{\lambda^2}$

   • A point on $L_1$ is $P_1(x_1, y_1, z_1) = (1, 2, -3)$.
   • The direction vector of $L_1$ is $\vec{d_1}(a_1, b_1, c_1) = (1, 2, \lambda^2)$.

   For Line 2 ($L_2$): $\frac{x - 3}{1} = \frac{y - 2}{\lambda^2} = \frac{z - 1}{2}$

   • A point on $L_2$ is $P_2(x_2, y_2, z_2) = (3, 2, 1)$.
   • The direction vector of $L_2$ is $\vec{d_2}(a_2, b_2, c_2) = (1, \lambda^2, 2)$.

   Step 2: Calculate the Vector Connecting the Two Points Next, we find the vector $\vec{P_1P_2}$ by subtracting the coordinates of $P_1$ from $P_2$. This vector is crucial for the coplanarity condition. $\vec{P_1P_2} = (x_2-x_1, y_2-y_1, z_2-z_1)$ $\vec{P_1P_2} = (3-1, 2-2, 1-(-3)) = (2, 0, 4)$

   Step 3: Apply the Coplanarity Condition Substitute the components of $\vec{P_1P_2}$, $\vec{d_1}$, and $\vec{d_2}$ into the determinant formula for coplanarity and set it to zero. $\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = 0$ $\begin{vmatrix} 2 & 0 & 4 \\ 1 & 2 & \lambda^2 \\ 1 & \lambda^2 & 2 \end{vmatrix} = 0$

   Step 4: Evaluate the Determinant and Form an Equation for $\lambda$ Expand the determinant along the first row to obtain an algebraic equation involving $\lambda$. $2 \cdot ( (2)(2) - (\lambda^2)(\lambda^2) ) - 0 \cdot ( (1)(2) - (\lambda^2)(1) ) + 4 \cdot ( (1)(\lambda^2) - (2)(1) ) = 0$ $2 \cdot (4 - \lambda^4) - 0 + 4 \cdot (\lambda^2 - 2) = 0$ $8 - 2\lambda^4 + 4\lambda^2 - 8 = 0$ $-2\lambda^4 + 4\lambda^2 = 0$

   Step 5: Solve the Equation for $\lambda$ Factor the equation to find the possible real values of $\lambda$. $-2\lambda^4 + 4\lambda^2 = 0$ Factor out $2\lambda^2$: $2\lambda^2 ( - \lambda^2 + 2 ) = 0$ This equation holds true if either $2\lambda^2 = 0$ or $(2 - \lambda^2) = 0$.

   • Case 1: $2\lambda^2 = 0$ $\lambda^2 = 0 \implies \lambda = 0$ This is one distinct real value.

   • Case 2: $2 - \lambda^2 = 0$ $\lambda^2 = 2 \implies \lambda = \pm\sqrt{2}$ These are two distinct real values: $\sqrt{2}$ and $-\sqrt{2}$.

   Step 6: Count the Distinct Real Values of $\lambda$ The distinct real values of $\lambda$ obtained are $0$, $\sqrt{2}$, and $-\sqrt{2}$. Thus, there are 3 distinct real values of $\lambda$ for which the lines are coplanar.

   Step 7: Verify the Solutions (Optional but Recommended)

   • For $\lambda = 0$: $L_1: \frac{x-1}{1} = \frac{y-2}{2} = \frac{z+3}{0}$. This implies $z+3=0 \implies z=-3$. $\vec{d_1}=(1,2,0)$. $L_2: \frac{x-3}{1} = \frac{y-2}{0} = \frac{z-1}{2}$. This implies $y-2=0 \implies y=2$. $\vec{d_2}=(1,0,2)$. Since $\vec{d_1}$ is not proportional to $\vec{d_2}$, the lines are not parallel. They intersect if coplanar. We can verify intersection by finding a common point. From $L_1$, $x=1+t, y=2+2t, z=-3$. From $L_2$, $x=3+s, y=2, z=1+2s$. Equating $y$ components: $2+2t=2 \implies t=0$. Substituting $t=0$ into $L_1$ gives point $(1,2,-3)$. Substituting $y=2$ into $L_2$ implies $\frac{x-3}{1} = \frac{z-1}{2}$. If $(1,2,-3)$ lies on $L_2$: $\frac{1-3}{1} = \frac{-3-1}{2} \implies -2 = -2$. This is consistent. So, $\lambda=0$ makes the lines intersecting and thus coplanar.
   • For $\lambda = \pm\sqrt{2}$: In both cases, $\lambda^2 = 2$. $\vec{d_1}=(1,2,2)$ and $\vec{d_2}=(1,2,2)$. Since $\vec{d_1} = \vec{d_2}$, the lines are parallel. To check if they are distinct or identical, we see if $P_1(1,2,-3)$ lies on $L_2$. Substitute $P_1(1,2,-3)$ into $L_2: \frac{x-3}{1} = \frac{y-2}{2} = \frac{z-1}{2}$. $\frac{1-3}{1} = \frac{2-2}{2} = \frac{-3-1}{2}$ $-2 = 0 = -2$. This is false because $0 \ne -2$. Thus, the lines are distinct and parallel for $\lambda = \pm\sqrt{2}$. Distinct parallel lines are always coplanar. All three values of $\lambda$ lead to coplanar lines.

3. Common Mistakes & Tips

   • Determinant Expansion Errors: Be extremely careful with signs and calculations when expanding the $3 \times 3$ determinant. A small arithmetic mistake can lead to incorrect values of $\lambda$.
   • Handling Zero Denominators: If a parameter can make a denominator zero (e.g., $\lambda^2=0$), remember that in the context of symmetric form of a line, $\frac{X}{0}$ implies $X=0$. This means the line lies in a plane parallel to a coordinate plane, and the corresponding direction vector component is zero. The determinant method generally accounts for this.
   • Counting Distinct Values: After solving the algebraic equation, ensure you count only the distinct real values. For example, $\lambda^2=0$ yields only one distinct value ($\lambda=0$), not two.
   • Parallel Lines: The determinant condition for coplanarity inherently covers both intersecting and parallel lines. If the lines are parallel, their direction vectors are proportional, making two rows of the determinant proportional, which automatically makes the determinant zero.

4. Summary

   To find the values of $\lambda$ for which two given lines are coplanar, we first identify a point and the direction vector for each line. Then, we form a $3 \times 3$ determinant using the components of the vector connecting the two points and the two direction vectors. Setting this determinant to zero gives an algebraic equation for $\lambda$. Solving this equation for distinct real values of $\lambda$ provides the answer. In this problem, the equation obtained was $2\lambda^2(2-\lambda^2)=0$, which yielded three distinct real values for $\lambda$: $0$, $\sqrt{2}$, and $-\sqrt{2}$.

5. Final Answer

The final answer is $\boxed{3}$, which corresponds to option (D).