The number of distinct real values of $$\lambda$$, for which the lines $$\frac{x-1}{1} = \frac{y-2}{2} = \frac{z+3}{\lambda^2}$$ and $$\frac{x-3}{1} = \frac{y-2}{\lambda^2} = \frac{z-1}{2}$$, are coplanar is

We begin by writing both lines in vector form. For the first line we have

$$\frac{x-1}{1}=\frac{y-2}{2}=\frac{z+3}{\lambda^{2}} = r$$

so that a general point on this line can be written as

$$\vec{r}_1=\begin{pmatrix}1\\2\\-3\end{pmatrix}+r\begin{pmatrix}1\\2\\\lambda^{2}\end{pmatrix},$$

where the point $$P_1(1,2,-3)$$ lies on the line and the direction vector is

$$\vec{a}=\begin{pmatrix}1\\2\\\lambda^{2}\end{pmatrix}.$$

For the second line we write

$$\frac{x-3}{1}=\frac{y-2}{\lambda^{2}}=\frac{z-1}{2}=s,$$

giving the vector form

$$\vec{r}_2=\begin{pmatrix}3\\2\\1\end{pmatrix}+s\begin{pmatrix}1\\\lambda^{2}\\2\end{pmatrix},$$

with point $$P_2(3,2,1)$$ and direction vector

$$\vec{b}=\begin{pmatrix}1\\\lambda^{2}\\2\end{pmatrix}.$$

Two lines in space are coplanar if and only if they are either parallel or intersecting. Both conditions are tested conveniently by the scalar triple product. The standard result is:

$$ (P_2-P_1)\cdot(\vec{a}\times\vec{b})=0 \quad\Longleftrightarrow\quad \text{lines are coplanar}. $$

(If $$\vec{a}\times\vec{b}=0$$ the lines are parallel; otherwise the same equation ensures intersection.)

First we compute the connecting vector

$$\vec{m}=P_2-P_1=\begin{pmatrix}3-1\\2-2\\1-(-3)\end{pmatrix}=\begin{pmatrix}2\\0\\4\end{pmatrix}.$$

Next we find the cross product $$\vec{a}\times\vec{b}$$:

$$ \vec{a}\times\vec{b}= \begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\ 1&2&\lambda^{2}\\ 1&\lambda^{2}&2 \end{vmatrix} = \mathbf{i}(2\cdot2-\lambda^{2}\lambda^{2})-\mathbf{j}(1\cdot2-\lambda^{2}\cdot1)+\mathbf{k}(1\cdot\lambda^{2}-2\cdot1). $$

Simplifying term by term, we get

$$ \vec{a}\times\vec{b} =\begin{pmatrix}4-\lambda^{4}\\-(2-\lambda^{2})\\\lambda^{2}-2\end{pmatrix} =\begin{pmatrix}4-\lambda^{4}\\\lambda^{2}-2\\\lambda^{2}-2\end{pmatrix}. $$

Now we form the scalar triple product:

$$ \vec{m}\cdot(\vec{a}\times\vec{b}) =\begin{pmatrix}2\\0\\4\end{pmatrix}\!\cdot\!\begin{pmatrix}4-\lambda^{4}\\\lambda^{2}-2\\\lambda^{2}-2\end{pmatrix} =2(4-\lambda^{4})+0(\lambda^{2}-2)+4(\lambda^{2}-2). $$

Multiplying out,

$$ 2(4-\lambda^{4})+4(\lambda^{2}-2) =8-2\lambda^{4}+4\lambda^{2}-8 =4\lambda^{2}-2\lambda^{4}. $$

Setting this equal to zero for coplanarity,

$$ 4\lambda^{2}-2\lambda^{4}=0 \;\Longrightarrow\; 2\lambda^{2}(2-\lambda^{2})=0. $$

Hence either

$$\lambda^{2}=0 \quad\text{or}\quad \lambda^{2}=2.$$

Case 1: $$\lambda^{2}=0 \Rightarrow \lambda=0.$$ Here $$\vec{a}\times\vec{b}\neq0,$$ so the lines intersect; they are certainly coplanar.

Case 2: $$\lambda^{2}=2 \Rightarrow \lambda=\sqrt{2}\; \text{or}\; \lambda=-\sqrt{2}.$$ For these values we have $$\vec{a}\times\vec{b}=0,$$ i.e. the direction vectors are parallel. Two parallel lines always lie in a common plane, so they are also coplanar.

Collecting the distinct real values, we have

$$\lambda=0,\;\; \lambda=\sqrt{2},\;\; \lambda=-\sqrt{2},$$

that is, exactly three real numbers.

Hence, the correct answer is Option C.