Plane through a point and the line of intersection of other two planes 
Find the equation of the plane through $(-1,4,2)$ and containing the line of intersection of the planes $$4x-y+z-2=0 \\ 2x+y-2z-3=0$$

My answer comes out to be:
$$-9x-67y+104=51$$
While the answer provided on the answer sheet is:
$$4x-13y+21z=-14$$
Could you please check if the answer calculated by me is correct or the one provided on the answer sheet.
 A: Other hint
The equation of sheaf of planes passing through the intersection of planes
4x-y+z-2=0 and 2x+y-2z-3=0:
$\lambda_1(4x-y+z-2)+\lambda_2(2x+y-2z-3)=0$. The specific plane of sheaf of planes determine numbers $\lambda_1,\,\lambda_2$, which are not simultaneously = $0$.
The equation of the plane through (−1,4,2) --> coordinates put into equation of sheaf of planes:
$\displaystyle \lambda_1(-4-4+ 2-2)+\lambda_2(-2+4-4-3)=0\Rightarrow \frac{\lambda_1}{\lambda_2}=-\frac{5}{8}$
$\displaystyle -\frac{5}{8}(4x-y+z-2)+(2x+y-2z-3)=0\Rightarrow \cdots \Rightarrow 4x-13y+21z+14=0$
A: HINT:
$$\begin{cases}
4x-y+z-2=0\\
2x+y-2z-3=0
\end{cases}\Longleftrightarrow$$
$$\begin{cases}
z=-4x+y+2\\
2x+y-2z-3=0
\end{cases}\Longleftrightarrow$$
$$\begin{cases}
z=-4x+y+2\\
2x+y-2(-4x+y+2)-3=0
\end{cases}\Longleftrightarrow$$
$$\begin{cases}
z=-4x+y+2\\
10x-y-7=0
\end{cases}\Longleftrightarrow$$
$$\begin{cases}
z=-4x+y+2\\
y=10x-7
\end{cases}\Longleftrightarrow$$
$$\begin{cases}
z=-4x+(10x-7)+2\\
y=10x-7
\end{cases}\Longleftrightarrow$$
$$\begin{cases}
z=6x-5\\
y=10x-7
\end{cases}$$
A: Notice, the equation of the plane passing through the line of intersection of planes: $4x-y+z-2=0$ & $2x+y-2z-3=0$ is given as $$4x-y+z-2+\lambda(2x+y-2z-3)=0$$ 
$$(2\lambda+4)x+(\lambda-1)y+(1-2\lambda)z=(3\lambda+2)\tag 1$$
Since, the above plane passes through the given point $(-1, 4, 2)$, hence it will satisfy the equation (1) of the plane as follows 
$$(2\lambda+4)(-1)+(\lambda-1)(4)+(1-2\lambda)(2)=(3\lambda+2)$$ 
$$-5\lambda-8=0\implies \lambda=\frac{-8}{5}$$
Now, substituting the value of $\lambda$ in (1), we get the equation of the plane as follows 
$$\left(2\left(\frac{-8}{5}\right)+4\right)x+\left(\frac{-8}{5}+4\right)y+\left(1-2\left(\frac{-8}{5}\right)\right)z=\left(3\left(\frac{-8}{5}\right)+2\right)$$
$$\color{blue}{4x-13y+21z=-14}$$
