Find an equation for the plane containing the point (1, 1, 1) and the line (x, y, z) = (1 + 2t, 3 − t, t).

Step-by-step explanation:

The line can be written as

Since the line is contained in the plane, if we take any two points of this line they will lie in the plane.

Give values  for t =0 and 1

We get two points (x,y,z) as (1,3,0) and (3,2,1)

Now we have another non collinearpoint (1,1,1) (given)

The plane equation can be written using these 3 points using the determinant.

x-x1 y-y1 z-z1

x2-x1 y2-y2 z2-z1

x3-x1 y3-y1 z3-z1    =0

Substitute the values form the points.

We get

Expand the determinant as

(x-1)(-1+2)-(y-3)(2-0)+z(-4-0) =0

x-1-2y+6-4z =0

x-2y-4z+5 =0

i used to know how to do this in 7th but i forget easily

step-by-step explanation:

idk what the heck he answer is, i’m just putting this here so i finish my account