Consider $p(x,y,z,t) = e^t(ax^2 + by^2 + cz^2)$ and I seek the values of $a,b,c$ which make $p\approx0$ at the point $(x=r\cos\theta,y=r\sin\theta,m,t)$ for all $t\ge0$ where $0 < \theta \le 2\pi$. A solution will be $(a,b,c)=\{(m/r)^2,(m/r)^2,-1\}$. How could I have found this in Mathematica?

I thought if I can obtain MinValue and MaxValue of p to both be within a small numerical range, then p would be approx zero. So I tried this:

p = Exp[t]*(a*x^2 + b*y^2 + c*z^2);
points = {x->r*Cos[theta],y->r*Sin[theta],z->m};
Minimize[p /. points, 0 < theta <= 2*pi && t >= 0, {a,b,c}]

But Mathematica returns the last line back to me.

I also tried this to no avail, but answers were not found either:

Reduce[p /.points == 0 && 0 < theta <= 2*Pi, {r,theta,t}, Reals]

Last of all I also tried:

Solve[ForAll[{r,theta,t}, p /. points == 0 && 0 < theta < 2*Pi], {a,b,c}]