Two planes are parallel if their normal vectors are parallel. That is, their gradients are scalar multiples of one another.
Let's denote the hyperboloid as $f(x,y,z) = 9x^2-45y^2+5z^2-45$ and the plane as $g(x,y,z) = x+5y-2z=7$.
Taking the gradient of each of these,
$$
\nabla f = \left< 18x, -90y, 10z \right>
$$
$$
\nabla g = \left< 1, 5, -2 \right>
$$
This yields the following equations (setting the components of each gradient equal to one another):
$$
18x = 1
$$
$$
-90y = 5
$$
$$
10z = -2
$$
Solving these gives the point $\left(\frac{1}{18}, -\frac{1}{18}, -\frac{1}{5}\right)$.