Answer
The equation of the plane is $17x-21y-13z=-28$.
Work Step by Step
Let $A$ and $B$ denote two points in the line ${\bf{r}}\left( t \right) = \left( {1,4, - 3} \right) + t\left( {2,1,1} \right)$ corresponding to $t=0$ and $t=1$, respectively. Thus,
$A = {\bf{r}}\left( 0 \right) = \left( {1,4, - 3} \right) + 0\left( {2,1,1} \right) = \left( {1,4, - 3} \right)$
$B = {\bf{r}}\left( 1 \right) = \left( {1,4, - 3} \right) + 1\left( {2,1,1} \right) = \left( {3,5, - 2} \right)$
Since the plane contains the line, so the points $A$ and $B$ lie in the plane. If the plane passes through $P = \left( {4, - 1,9} \right)$, so $P$ also lies in the plane. Thus, we have the vectors:
$\overrightarrow {AB} = B - A = \left( {3,5, - 2} \right) - \left( {1,4, - 3} \right) = \left( {2,1,1} \right)$
$\overrightarrow {AP} = P - A = \left( {4, - 1,9} \right) - \left( {1,4, - 3} \right) = \left( {3, - 5,12} \right)$
Since $\overrightarrow {AB} $ and $\overrightarrow {AP} $ lie in the plane, their cross product ${\bf{n}} = \overrightarrow {AB} \times \overrightarrow {AP} $ is a vector normal to the plane.
${\bf{n}} = \overrightarrow {AB} \times \overrightarrow {AP} = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
2&1&1\\
3&{ - 5}&{12}
\end{array}} \right|$
${\bf{n}} = \left| {\begin{array}{*{20}{c}}
1&1\\
{ - 5}&{12}
\end{array}} \right|{\bf{i}} - \left| {\begin{array}{*{20}{c}}
2&1\\
3&{12}
\end{array}} \right|{\bf{j}} + \left| {\begin{array}{*{20}{c}}
2&1\\
3&{ - 5}
\end{array}} \right|{\bf{k}}$
${\bf{n}} = 17{\bf{i}} - 21{\bf{j}} - 13{\bf{k}}$ ${\ \ }$ or ${\ \ }$ ${\bf{n}} = \left( {17, - 21, - 13} \right)$
By Theorem 1 of Section 13.5, the equation of the plane is
$17x - 21y - 13z = d$,
where $d$ is to be determined.
We choose $P = \left( {4, - 1,9} \right)$ and substitute it in the equation above to find $d$:
$17\cdot4 - 21\cdot\left( { - 1} \right) - 13\cdot9 = d$
$d=-28$
Thus, the equation of the plane is $17x-21y-13z=-28$.
