Find the plane parallel to the line connecting point a (2,5, - 3) and point B (3, - 2,2) through the intersection of two planes π 1:2x + Y-Z + 1 = 0 and π 2: x + y + 2Z + 1 = 0 equation

Find the plane parallel to the line connecting point a (2,5, - 3) and point B (3, - 2,2) through the intersection of two planes π 1:2x + Y-Z + 1 = 0 and π 2: x + y + 2Z + 1 = 0 equation

Let the equation of the plane be 2x + Y-Z + 1 + a (x + y + 2Z + 1) = 0
Simplification: (2 + a) x + (1 + a) y + (2a-1) Z + 1 + a = 0
Since AB is parallel to the plane, the vector AB is perpendicular to the normal vector of the plane, and the vector AB = (1, - 7,5)
So: (2 + a) - 7 (1 + a) + 5 (2a-1) = 0, a = 2.5 is substituted into the plane equation
The result is: 9x + 7Y + 8Z + 7 = 0

The plane equation of the straight line x + 5Y + Z = 0, x-z + 4 = 0 and the plane x-4y-8z + 12 = 0 is solved

x+20y+7z-12=0

A plane passes through a straight line {x + 5Y + Z = 0, x-z + 4 = 0}, and is perpendicular to the plane x-4y-8z + 12 = 0. The plane equation is solved

Let the square equation be x + 5Y + Z + a (x-z + 4) = 0 (1 + a) x + 5Y + (1-A) Z + 4A = 0, because the plane is perpendicular to the known plane x-4y-8z + 12 = 0, so their normal vectors are perpendicular, that is, 1 × (1 + a) - 4 × 5-8 × (1-A) = 01 + a-20-8 + 8A = 09A = 27a = 3, so the equation is: (1 + 3) x + 5Y + (1-3) Z + 4 × 3 = 0, that is 4x + 5y-2z +

X square + y square - 4Y + 8y + 20 = 0, the value of X + y can be obtained by using the matching method flexibly

x^2 y^2-4x 8y 20=0
x^2-4x 4 y^2 8y 16=0
(x-2)^2 (y 4)^2=0
x=2 y=-4
x y=-2