High number, vector, plane. Find the bisector equation of two dihedral angles formed by plane 2x-y + Z = 5 and plane x + 2Y + Z = 9

High number, vector, plane. Find the bisector equation of two dihedral angles formed by plane 2x-y + Z = 5 and plane x + 2Y + Z = 9

An advanced number problem: finding the equation of plane B with a distance of 5 and plane a: 3x + 2y-2 √ 3z-2 = 0

Let the plane be 3x + 2y-2 √ 3Z + M = 0
If the distance between the two planes is d = | m + 2 | / 5 = 5, then M = 23 or M = - 27,
So the plane obtained is 3x + 2y-2 √ 3z-27 = 0 or 3x + 2y-2 √ 3Z + 23 = 0

Find the equation of the line passing through point P (1,0, - 2) and parallel to plane 3x-y + 2z-1 = 0 and intersecting with the line (x-1) / 4 = (Y-3) / - 2 = Z / 1
The answer is 3x-y + 2Z + 1 = 0 and - 7x + 8y + 12z + 31 = 0

The two equations in the answer are simultaneous
First of all, we can determine the first equation in the answer. Because P is not in 3x-y + 2z-1 = 0, the line passing through P and parallel to the plane must be in the plane passing through P and parallel to the plane. The equation of the plane is 3x-y + 2Z + 1 = 0
Then the line intersects with (x-1) / 4 = (Y-3) / - 2 = Z / 1, and it is known that the line is in the plane specified by the first equation, (x-1) / 4 = (Y-3) / - 2 = Z / 1 intersects with 3x-y + 2Z + 1 = 0 to get the intersection point Q (calculated by myself), and PQ two points determine the line

Find the linear equation which is parallel to two planes x + 2Z = 1 and y-3z = 2 through point P (- 1,2,3)

If it is parallel to two planes, it must be parallel to the intersection of two planes
Let z = t
be
x=-2t+1
y=3t+2
z=t
So the normal vector of the intersection line is (- 2,3,1)
Therefore, the equation of a straight line passing through P and parallel to the intersection line is as follows:
(x+1)/(-2)=(y-2)/2=(z-3)/3