Two degree equation of mutually perpendicular 3D planes

Question

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

My approach is as follow:

Let the two plane equation be $ax+by+cz=0$ &$a'x+b'y+c'z=0$

As they are mutually perpendicular then $aa'+bb'+cc'=0$

On multiplying both plane we get $aa'x^2+bb'y^2+cc'z^2+(ab'+a'b)xy+(bc'+b'c)yz+(ac'+a'c)xz=0$

On comparing we get $~p+q=-1~$ after this step, I am not able to approach further.

Answer 1

Homogeneous equation of second degree in $~x,~y~$ and $~z~$, $$px^2+y^2+qz^2+2yz+zx+3xy=0$$ represents a pair of perpendicular plains if $$\begin{vmatrix} p & \dfrac 32 & \dfrac 12 \\ \dfrac 32 & 1 & 1 \\ \dfrac 12 & 1 & q \end{vmatrix}=0\qquad\text{and}\qquad p+1+q=0$$ $$\implies 4pq-4p-9q+5=0\qquad\text{and}\qquad p=-1-q$$ $$\implies 4q(-1-q)-4(-1-q)-9q+5=0$$ $$\implies q=-3,~~~~\dfrac 34$$ and hence $~p=2,~~-\dfrac 74~.$

Therefore $~p-q~=5,~~-\dfrac 52~.$

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Note: A homogeneous equation of second degree in $~x,~y~$ and $~z~$, $$ax^2+by^2+cz^2+2fyz+2gzx+2hyx=0$$ represents a pair of perpendicular plains if $$\begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix}=0\qquad\text{and}\qquad a+b+c=0$$