Where does the line through (1, 0, 1) and (4, −3, 3) intersect the plane x + y + z = 6?

You can use the equation of a line passing through 2 points in three dimensions.

The considered line will intersect the given plane on (x,y,z) = (7,-6,5)

Suppose the points be (a,b,c) and (d,e,f)

Then we have;

and the equation of the line is given as

Since the two points through which the line passes are (1,0,1) and (4,-3,3)

Thus, we have

And thus, equation of line is

Since this line passes through given plane x + y + z = 6, thus,

we have one common point between these two structures.

using first two equality and then first and third terms from the equation of the line to get y and z in terms of x

Putting these values in equation of plane, we get:

Thus, from y and z's expression in terms of x, we get
y = 1-x = -6

and z = (2x +1)/3 = 5

Thus, the point of intersection of considered line and given plane is

(x,y,z) = (7,-6,5)

Learn more about intersection of line and plane here:

link

Parametrically for real t the line is

That's three equations; we substitute into

The intersection is a single point (7,-6,5).

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