Find the equation of the plane passing through the point P(1,1,1,) and containing the line r=(-3i+j+5k) + A(3i-j-5k). Also show that the plane contains the line r=(-i+2j+5k) + B(i-2j-5k).

Equation of plane

A'(x - x₁) + B'(y - y₁) + C'(z - z₁) = 0

Given line
r = -3i + j + 5k + A(3 i - j -5k)

Taking A = 1

we get r = 0, thus (0, 0, 0) is a point on the line

Taking  A = 2

We get r = 3i - j - 5k,  hence (3, -1, -5) is also a point on the line

Hence the plane passes through three points (0,0,0) , (1, 1, 1) and (3, -1,-5)
As A'(x₁ -x₂ ) +B'(y₁-y₂) +C'(z₁-z₂) = 0 is a equation of the plane passing through two points
So

A'(1 - 0) + B'(1 - 0) + C'(1 - 0) = 0
A' +B'+C' = 0 (1)

A'(3 - 0) + B'(-1 - 0) + C'(-5 - 0) = 0

3A' - B' - 5C' = 0 (2)

A'/ (-5 + 1) = -B'/(-5 - 3) = C'/(-1 - 3)

A'/-4 = B'/8 = C'/-4

(A':B':C') = (1: -2: 1)

Hence the equation of the given  plane is

x - 2y + z = 0  is the required equation of the plane

its normal vector is n = (i - 2j + k)

To prove that the plane contains the second line r₂ = (B - 1)i + (2 - 2B)j + (5 - 5B)k

Take its scalar product with the normal of the given plane i - 2j + k .

So r₂.n = {(B - 1)i + (2 - 2B)j + (5 - 5B)k}.( i - 2j + k )
= B - 1 - 4 + 4B + 5 - 5B
= 0

The scalar product is zero which ensures the second line is parallel to the plane.

Now take B = 1, we get r₂ = 0 which means that the line passes through origin as well.

Hence the second line is parallel to the plane and passes through a point in the plane which ensures that the line is contained in the plane.