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 $x+y+z=0$, 
$$A+B+C=0$$
 A: When the equation of a plane is written as
$$ ax +by+cz+d = 0 $$
Then you know that the vector $\left(a\quad b\quad c\right)$ stands perpendicular on that plane, i.e. a normal of the plane. Remind yourself that a normal can be multiplied with a real scaling factor and still be a normal: 
$$ \lambda ax + \lambda by+\lambda cz+ \lambda d = 0 $$
This is why the OP only has 2 equations and 3 unknowns. The scaling factor is an extra degree of freedom. So the OP already presents a correct solution, all that needs to be done is work it out.
When you think about the problem geometrically, it is much easier:
Since you search for a plane which is perpendicular to $9x−7y+6z+48=0$ and $x+y+z=0$, you know that it must be parallel to the plane formed by the two normals of those planes, i.e. $\left(9\quad -7 \quad 6\right)$ and $\left(1\quad 1\quad 1 \right)$. The normal of this plane will, therefore, be the cross-product of those two vectors and directly an answer to the OP's $A$, $B$ and $C$.
A: 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}
