Can a plane be drawn through the lines $\overrightarrow{r} = \left(\hat{i} + 2 \hat{j} - 4 \hat{k}\right) + t \left(2 \hat{i} + 3 \hat{j} + 6 \hat{k}\right)$ and $\overrightarrow{r} = \left(3 \hat{i} + 3 \hat{j} - 5 \hat{k}\right) + s \left(-2 \hat{i} + 3 \hat{j} + 8 \hat{k}\right)$?
The equations of the given lines are
$L_1: \;\; \overrightarrow{r} = \left(\hat{i} + 2 \hat{j} - 4 \hat{k}\right) + t \left(2 \hat{i} + 3 \hat{j} + 6 \hat{k}\right)$
$L_2: \;\; \overrightarrow{r} = \left(3 \hat{i} + 3 \hat{j} - 5 \hat{k}\right) + s \left(-2 \hat{i} + 3 \hat{j} + 8 \hat{k}\right)$
Lines $L_1$ and $L_2$ are of the form
$\overrightarrow{r} = \overrightarrow{a_1} + t \overrightarrow{b_1}$ $\;$ and $\;$ $\overrightarrow{r}= \overrightarrow{a_2} + s \overrightarrow{b_2}$ $\;$ respectively.
Here,
$\overrightarrow{a_1} = \hat{i} + 2 \hat{j} - 4 \hat{k}$, $\;$ $\overrightarrow{b_1} = 2 \hat{i} + 3 \hat{j} + 6 \hat{k}$
$\overrightarrow{a_2} = 3 \hat{i} + 3 \hat{j} - 5 \hat{k}$, $\;$ $\overrightarrow{b_2} = -2 \hat{i} + 3 \hat{j} + 8 \hat{k}$
Now,
$\begin{aligned}
\overrightarrow{a_2} - \overrightarrow{a_1} & = \left(3 \hat{i} + 3 \hat{j} - 5 \hat{k}\right) - \left(\hat{i} + 2 \hat{j} - 4 \hat{k}\right) \\\\
& = 2 \hat{i} + \hat{j} - \hat{k}
\end{aligned}$
$\begin{aligned}
\overrightarrow{b_1} \times \overrightarrow{b_2} & = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
2 & 3 & 6 \\
-2 & 3 & 8
\end{vmatrix} \\\\
& = \hat{i} \left(24 - 18\right) - \hat{j} \left(16 + 12\right) + \hat{k} \left(6 + 6\right) \\\\
& = 6 \hat{i} - 28 \hat{j} + 12 \hat{k}
\end{aligned}$
$\begin{aligned}
\left(\overrightarrow{a_2} - \overrightarrow{a_1}\right) \cdot \left(\overrightarrow{b_1} \times \overrightarrow{b_2}\right) & = \left(2 \hat{i} + \hat{j} - \hat{k}\right) \cdot \left(6 \hat{i} - 28 \hat{j} + 12 \hat{k}\right) \\\\
& = 12 - 28 - 12 = -28 \neq 0
\end{aligned}$
$\because$ $\;$ $\left(\overrightarrow{a_2} - \overrightarrow{a_1}\right) \cdot \left(\overrightarrow{b_1} \times \overrightarrow{b_2}\right) \neq 0$ $\implies$ lines $L_1$ and $L_2$ are not coplanar.
$\because$ $\;$ The given lines are not coplanar, a plane cannot be drawn through the lines $L_1$ and $L_2$.