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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 $x+y+z=0$, $$A+B+C=0$$

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    $\begingroup$ And what is the problem? You have the two equations you need, don't you? Just solve them. $\endgroup$ – Arthur May 6 at 6:33
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    $\begingroup$ Hint: if planes are perpendicular, then so are their normals. $\endgroup$ – amd May 6 at 8:23
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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 $C$.

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  • $\begingroup$ You say "The normal of this plane", I would argue it's "A normal of this plane". Also, he's not really missing any information. The problem is completely solvable from what the OP has already done. See my comment to the question post itself above. $\endgroup$ – Arthur May 6 at 7:42
  • $\begingroup$ @Arthur you are completely correct. I have updated the answer. $\endgroup$ – kvantour May 6 at 7:54
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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}

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