Answer
\[
14=-4 x+13 y-21 z
\]
Work Step by Step
From the equations of the given planes, we get that
\[
\begin{aligned}
y &=-2 x+2 z+3=4 x+z-2 \\
\Rightarrow z &=2 x+4 x-3-2=-5+6 x
\end{aligned}
\]
So, for $t=x$ we get that
\[
t=x, 6 t+4 t-2-5=y, -5+6 t=z
\]
\[
\Rightarrow t=x, \quad -5+6 t=z , \quad -7+10 t=y
\]
Since the plane $d=a x+b y+c z$ contains the line, the normal vector to the plane $\vec{n}=\langle a, b, c\rangle$ is orthogonal to $\vec{v}=\langle 1,10,6\rangle$ (parallel line vector ). And then
\[
\begin{aligned}
0=\vec{n} \cdot \vec{v} & \Rightarrow 0=\langle a, b, c\rangle\langle 1,10,6\rangle \\
& \Rightarrow 0=a+10 b+6 c \\
& \Rightarrow -a=10 b+6 c
\end{aligned}
\]
For $0=t$, we get that
\[
-5=z, \quad -7=y \quad, \quad x=0
\]
So the plane and the line contains the point $Q(0,-7,-5) .$ This implies that
\[
\begin{aligned}
d=a(0)+b(-7)+c(-5) & \\
\Rightarrow d=\quad-7 b-5 c
\end{aligned}
\]
Since the plane passes through $(-1,4,2)P$, we get that
\[
d=-a+4 b+2 c
\]
We have found the following system of linear equations
\[
\begin{array}{l}
-6 c-10 b=a \\
2 c-d+4 b=a \\
-5 c-7 b=d
\end{array}
\]
The solution to the linear equations is:
\[
\frac{13 d}{14}=b, -\frac{3 d}{2}=c, \quad -\frac{2 d}{7}=a
\]
So the equation of the plane is
\[
\begin{array}{l}
d=\frac{13 d}{14} y-\frac{3 d}{2} z -2 \frac{d}{7} x\\
\Rightarrow14=13 y-21 z-4 x
\end{array}
\]
