Let's first determine the intersection of the planes 2x-y=0 and y+z=-1. We determine normal vectors from those two planes (why this works is outside the scope of this answer).
n1 = [2,1,0]
n2 = [0,1,1]
where n1 and n2 represent normal vectors of the planes. A line that exists as the intersection of two planes must be perpendicular to both planes. Thus, we can determine a vector (v1) parallel to the intersection of the planes by taking a cross product of n1 and n2.
v1 = n1 x n2 = [1,-2,2]
The final plane must be perpendicular to both v1 and a vector parallel to the line x=3t, y=2+t, z=-2t, which will be called v2. We first determine a simple choice of v2.
v2 = [3,1,-2]
So, now we can find a possible n3, a normal of the final plane, by taking a cross product of v1 and v2.
n3 = v1 x v2 = [2,8,7]
We will also need a point in the plane. A simple choice is picked from t=0, giving p=(0,2,0). Thus, we can say that the equation of the plane is:
2x + 8(y-2) + 7z = 0
2x + 8y + 7z = 16