Answer
$33x+10y+4z=190$
Work Step by Step
We know that the plane contains the point $P_0 = (6,0,-2)$ and the line defined by $x = 4 -2t, y = 3 +5t, z = 7 +4t$. So, we must find a normal vector that is orthogonal to two vectors on the plane.
1. Rewrite the equation of the line in vector form: $L = \langle{4 , 3, 7}\rangle + t\langle{-2, 5, 4}\rangle$.
2. We can find two points on the line by setting $t =0$, $t = 1$. We have $P_1 = (4, 3, 7)$ and $P_2 = (2, 8, 11)$. The corresponding position vectors are $\vec{P_1} = \langle{4, 3, 7}\rangle$ and $\vec{P_2} = \langle{2, 8, 11}\rangle$.
3. Now, since the plane contains the line $L$, it also contains $P_1$ and $P_2$. Hence we can find two vectors on the plane with $\vec{v_1} = \vec{P_0P_1}$ and $\vec{v_2} = \vec{P_0P_2}$.
4. Using vector subtraction, we find that $\vec{v_1} = \langle{-2, 3, 9}\rangle$ and $\vec{v_2} = \langle{-4, 8, 13}\rangle$.
5. To get the normal vector, take the cross product of $\vec{v_1}$ and $\vec{v_2}$: $\vec{n} = \vec{v_1} \times \vec{v_2} = \langle{-33, -10, 4}\rangle$.
6. Write the equation of the plane in scalar form with normal vector $\vec{n}$ and point $P_0$:
\begin{equation}
-33(x - 6) - 10(y -0) -4(z +2) = 0
\end{equation}
7. Simplify the equation of the plane to get our final answer:
\begin{equation}
33x + 10y + 4z = 190
\end{equation}
