Answer
The equation of the plane is $x-2y+z=0$.
Work Step by Step
1. Consider the line ${\bf{r}}\left( t \right) = \left( {t + 1,2t,3t - 1} \right)$.
We choose two points ${P_1}$ and ${Q_1}$ on the line corresponding to $t=0$ and $t=1$, respectively. So, we have ${P_1} = \left( {1,0, - 1} \right)$ and ${Q_1} = \left( {2,2,2} \right)$.
Write the vector ${\bf{u}} = \overrightarrow {{P_1}{Q_1}} = {Q_1} - {P_1} = \left( {1,2,3} \right)$. Since the plane contains the line ${\bf{r}}$, it also contains the vector ${\bf{u}}$.
2. Consider the point $P = \left( { - 1,0,1} \right)$.
Since $P$ and ${P_1}$ lie in the plane. We write the vector ${\bf{v}}$ as
${\bf{v}} = \overrightarrow {{P_1}P} = P - {P_1} = \left( { - 1,0,1} \right) - \left( {1,0, - 1} \right)$
${\bf{v}} = \left( { - 2,0,2} \right)$
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
${\bf{n}} = {\bf{u}} \times {\bf{v}} = \left( {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
1&2&3\\
{ - 2}&0&2
\end{array}} \right)$
${\bf{n}} = \left| {\begin{array}{*{20}{c}}
2&3\\
0&2
\end{array}} \right|{\bf{i}} - \left| {\begin{array}{*{20}{c}}
1&3\\
{ - 2}&2
\end{array}} \right|{\bf{j}} + \left| {\begin{array}{*{20}{c}}
1&2\\
{ - 2}&0
\end{array}} \right|{\bf{k}}$
${\bf{n}} = 4{\bf{i}} - 8{\bf{j}} + 4{\bf{k}}$
By Eq. (5) of Theorem 1, the equation of the plane is of the form
(1) ${\ \ \ }$ $4x - 8y + 4z = d$,
where $d$ is to be determined.
Next, we choose a point on the plane, say $P = \left( { - 1,0,1} \right)$ which satisfies equation (1). Substituting it in equation (1) gives $d=0$. Hence, the equation of the plane is $4x-8y+4z=0$. Simplification yields $x-2y+z=0$.
