Find an equation for the plane that contains the following line and passes through point P How do you determine the plane which contains the line
\begin{align}
x & = -1 + 3t \\
y & = 5 + 2t \\
z & = 2 + t
\end{align}
and passes through the point $P = (2,4,-1)$?

the line is
\begin{equation}
(-1,5,2) + t(3,2,1)
\end{equation}
Then the plane contains the segment
$\overline{(-1,5,2)(2,4,-1)}$ $\quad((-3,1,3) =  (-1,5,2) -((2,4,-1))$ and the plane $\Pi$ has normal $(3,2,1) \times (-3,1,3) = (5,-12,9)$. Hence the plane is $5x -12y + 9z + d = 0$. As $(2,4,-1) \in \Pi$, we have $d = 47$. Hence, the plane is
$5x -12y +9z +47=0$.
