Answer
The equation of the plane is $13x + y - 5z = 0$.
Work Step by Step
1. Consider the line ${{\bf{r}}_1}\left( t \right) = \left( {t,2t,3t} \right)$.
We choose two points ${P_1}$ and ${Q_1}$ on the line corresponding to $t=0$ and $t=1$, respectively. So, we have ${P_1} = \left( {0,0,0} \right)$ and ${Q_1} = \left( {1,2,3} \right)$.
Write the vector ${\bf{u}} = \overrightarrow {{P_1}{Q_1}} = {Q_1} - {P_1} = \left( {1,2,3} \right)$. Since the plane contains ${{\bf{r}}_1}$, it also contains the vector ${\bf{u}}$.
2. Consider the line ${{\bf{r}}_2}\left( t \right) = \left( {3t,t,8t} \right)$.
We choose two points ${P_2}$ and ${Q_2}$ on the line corresponding to $t=0$ and $t=1$, respectively. So, we have ${P_2} = \left( {0,0,0} \right)$ and ${Q_2} = \left( {3,1,8} \right)$.
Write the vector ${\bf{v}} = \overrightarrow {{P_2}{Q_2}} = {Q_2} - {P_2} = \left( {3,1,8} \right)$. Since the plane contains ${{\bf{r}}_2}$, it also contains the vector ${\bf{v}}$.
Since ${\bf{u}}$ and ${\bf{v}}$ lie in the plane, their cross product is normal to the plane. So, the vector normal to the plane ${\bf{n}}$ is
${\bf{n}} = {\bf{u}} \times {\bf{v}} = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
1&2&3\\
3&1&8
\end{array}} \right|$
${\bf{n}} = \left| {\begin{array}{*{20}{c}}
2&3\\
1&8
\end{array}} \right|{\bf{i}} - \left| {\begin{array}{*{20}{c}}
1&3\\
3&8
\end{array}} \right|{\bf{j}} + \left| {\begin{array}{*{20}{c}}
1&2\\
3&1
\end{array}} \right|{\bf{k}}$
${\bf{n}} = 13{\bf{i}} + {\bf{j}} - 5{\bf{k}}$
By Eq. (5) of Theorem 1, the equation of the plane is of the form
(1) ${\ \ \ }$ $13x + y - 5z = d$,
where $d$ is to be determined.
Now, the lines ${{\bf{r}}_1}$ and ${{\bf{r}}_2}$ pass through the origin corresponding to $t=0$. Since they lie in the plane, so the plane also passes through the origin. This means that $\left( {0,0,0} \right)$ satisfies equation (1). Substituting it in equation (1) we obtain $d=0$. Hence, the equation of the plane is $13x + y - 5z = 0$.
