How do I solve for $p-q$? Question: 

If the equation $px^2+y^2+qz^2+2yz+zx+3xy=0$ represents a pair of perpendicular planes,
  then $p-q=?$

My attempts:
I started off taking two planes:
$$Ax+By+Cz+D=0 \qquad \text{and} \qquad A^{'}x+B^{'}y+C^{'}z+D^{'}=0, $$
where $AA^{'}+BB^{'}+CC{'}=0$ (Due to inter perpendicularity of normals of the plane). I am  basically trying to figure out some correlation between pair of straight lines and pair of planes but No luck yet.
And by looks of my method, it seems that it's gonna be a long hour solving this. Can someone please suggest a better method?  
Cheers!
 A: I suspect from the way this problem is worded is that there is some  slick and clever for finding $p-q$; perhaps someone else knows that.
But I can think of a straightforward method for calculating $p$ and $q$, employing the concept of a quadratic form combined with some geometric thinking (I'll throw in that the value of $p+q$ comes out very nicely, as you'll see; but I cannot think of any reason for a nice value of $p-q$). 
The idea for solving this problem is to think of your original equation as a quadratic form, and then to put the quadratic form into a better format by a change of basis matrix. This way, you will spend your hour honing your knowledge of quadratic forms rather than doing a long boring calculation (not that you can entirely avoid the calculation part).
You can rewrite the equation in the form
$$v \cdot M \cdot v^T = 0
$$
where
$$v = \langle x,y,z \rangle
$$
and
$$M = \begin{pmatrix} 
p & \frac{3}{2} & \frac{1}{2} \\
\frac{3}{2} & 1 & 1 \\
\frac{1}{2} & 1 & q
\end{pmatrix}
$$
Now for some geometry. You are given that the original equation is the equation of two perpendicular planes, call them $P_1,P_2$. Clearly $(0,0,0)$ is a solution of your given equations. Also, for any solution $(x,y,z)$ clearly $(rx,ry,rz)$ is still a solution no matter the value of $r$. It follows that both of the planes $P_1,P_2$ must pass through the point $(0,0,0)$. We may therefore rotate Euclidean space holding the origin fixed so that $P_1$ goes to the $x',y'$ plane and $P_2$ goes to the $y',z'$ plane. This rotation may be realized by a $3 \times 3$ orthogonal matrix $Q$ which transforms $(x,y,z)$ coordinates to $(x',y',z')$ coordinates.  When we carry out this coordinate change, in the new coordinate system we obtain a symmetric matrix which I will write in the form
$$Q M Q^{-1} = Q M Q^T = \begin{pmatrix}
a & \frac{b}{2} & \frac{c}{2} \\
\frac{b}{2} & d & \frac{e}{2} \\
\frac{c}{2} & \frac{e}{2} & f
\end{pmatrix}
$$
This defines a new equation
$$a \, x'{}^2 + d \, y'{}^2 + f \, z'{}^2 + b  \, x'{}y'{} + c  \, x'{}z'{} + e  \, y'{}z'{} = 0
$$
whose solution set is the union of the $x',y'$ and $y',z'$ planes. Up to scalar multiple, that equation must be the same as $x' z' = 0$ and so $a=b=d=e=f=0$ and we have
$$Q M Q^{-1} = Q M Q^T =\begin{pmatrix}
0 & 0 & \frac{c}{2} \\
0 & 0 & 0 \\
\frac{c}{2} & 0 & 0
\end{pmatrix} = M'
$$
Since the characteristic polynomial of a matrix is invariant under conjugation, we conclude that the characteristic polynomial of your matrix must equal the characteristic polynomial of $M'$, i.e.
\begin{align*}
det(M-tI) &= det(M'-tI) \\ &= -t^3 + \frac{c^2}{4} t
\end{align*}
This gives you some tractable equations in $p$ and $q$ to play with. For instance, the $t^2$ coefficient equals the trace which equals zero and which is invariant under conjugation, hence $p+1+q=0$ (if only you were asked for the value of $p+q$ you'd be done). 
The constant coefficient is the determinant of $M'$ which is zero and hence the determinant of $M$ is zero, from which I get
$$pq-p-\frac{9}{4}q+\frac{5}{4}=0
$$
And then there's that $t$ coefficient whose name I don't know, from which I get the equation
$$pq + p + q - \frac{15}{4} = \frac{c^2}{4}
$$
The rest is not-bad-at-all algebra...
