# Finding the values of $a$ and $b$ so that the plane $ax+by+2z+1 = 0$ intersect two other planes in a line

I working on this problem: Find the values of $a$ and $b$ so that the plane $P_3 = ax+by+2z+1 = 0$ intersect the two other planes:
$$P_2 = x-2y+z+5 = 0,P_1 = 3x-3y-2z-1 = 0$$ in a line and describe the intersection line in parameter form.

My work:

We first try to find the line of intersection of the given planes $P_1, P_2$:

$x-2y+z+5 = 0$
$3x-3y-2z-1 = 0$

with elimination this gives us:

$3y -5z -16 = 0$ and with $z = t$

$x = -\frac{17}{3} + \frac{7t}{3}, y = \frac{16}{3} + \frac{5t}{3}, z = t$

Which is the parametric equation of the intersection line of the planes $P_1, P_2$. Now we need to find the values for $a, b$ that makes the plane $ax+by+2z+1 = 0$ parallell to the line.

This happens when the normal vector and the gradient vector are orthogonal:

$$(a, b, 2) \cdot (7,5,3) = 0 \rightarrow -7a = 5b +6$$

From this i can guess the right answer $a = -3, b = 3$ but obviously I need more information to solve this properly, what else can I do to work this out?

The plane should not only be parallel to the line, it should contain the line as well. For this, find an arbitrary point on the line (for example, $t = 2 \rightarrow (-11, 12, 2)$ and substitute it in the equation of the plane ($-11a + 12b + 2(2) + 1 = 0$). Now you have two linear equations which you can solve to find $a$ and $b$.