Consider the line through the points (3,2,5) and (1,1,1). Consider the plane A that is perpendicular to this line, and passing through the point (-1,0,2). Consider the plane B that passes through the points (1,-1,0), (0,2,0), and (0,5,2). Consider the intersection of plane A and plane B. Write the equation of the line of intersection.

Answer 1

A lot of considering taking place here.

plane A:
direction of given first line = (2,1,4)
so that becomes the normal to the plane and its
equation must be 2x + y + 4z = C
but it passes through (-1,0,2), so
-2 + 0 + 8 = C = 6
PLane A: 2x + y + 4z = 6

Plane B:
two directions are (1,-3,0) and (0,3,2)
the cross product is (6,2,-3) , which is the direction of the normal, so
6x + 2y - 3z = K
(1,-1,0) is on it, thus 6 - 2 + 0 = k = 4
Plane B : 6x + 2y -3z = 4

double equation of plane A
4x + 2y + 8z = 12
equation of plane B
6x + 2y - 3z = 4

subtract them:
2x - 11z = -8
x = (11z -8)/2

let z = 0, x = -4,
in plane A:
-8 + y + 0 = 6
y = 14 -----> point on line is (-4, 14, 0)

let z = 2, x = 7
in plane A:
14 + y +8 = 6
y = -16 -----> another point on the line is (7, -16, 2)

Now you have two points on your line.
Use whichever method you use to find the line

Answer 2

Thank you so much!