Find the equation of the plane through the point $(2,1,4)$ Find the equation of the plane through the point $(2,1,4)$ and perpendicular to each of the planes $9x-7y+6z+48=0$ and $x+y+z=0$.
My attempt:
The equation of the plane passing through the point $(2,1,4)$ is given by 
$$A(x-2)+B(y-1)+C(z-4)=0$$
Here, $A,B,C$ represents the direction ratios of normal to the plane. 
Since, it is perpendicular to the plane $9x-7y+6z+48=0$, 
$$9A-7B+6C=0$$
Also, the plane is perpendicular to another plane $x+y+z=0$, 
$$A+B+C=0$$

By solving the second and third equations, we have
\begin{align}
B&=\frac{3}{13}A\\
C&=-\frac{16}{13}A
\end{align}
So the plane is
\begin{align}
A(x-2)+3(y-1)A/13-16(z-4)A/13 &=0\\
13(x-2)+3(y-1)-16(z-4) &=0
\end{align}
