equation of a plane?

Question

find an equaition of the plane that passes through the point
(8,-3,4) and contains the line with symmetric equations
x=5y=7z

Answer 1

A line and a point not on the line are sufficient to define a plane.

Given point P(8, -3, 4) and line x = 5y = 7z.

Rewrite the equation of the line.
t = x/35 = y/7 = z/5

The directional vector of the line, and also of the plane, is:
v = <35, 7, 5>

A point on the line is obviously O(0, 0, 0).

A second directional vector is:
u = OV = <8, -3, 4>

The normal vector n, to the plane is orthogonal to both directional vectors. Take the cross product.

n = u X v = <8, -3, 4> X <35, 7, 5> = <-43, 100, 161>

With the normal vector to the plane and a point on the plane we can write the equation of the plane. Let's use O(0, 0, 0)

-43(x - 0) + 100(y - 0) + 161(z - 0) = 0
-43x + 100y + 161z = 0