Parametrically for real t the line is
[tex](x,y,z) = (1-t)(1,0,1) + t(4,-3,3)[/tex]
[tex](x,y,z) = (1,0,1) + t(4-1,-3-0,3-1)[/tex]
[tex](x,y,z) = (1,0,1) + t(3,-3,2)[/tex]
That's three equations; we substitute into
[tex]x+y+z=6[/tex]
[tex](1+3t)+(0-3t) + ( 1+2t)=6[/tex]
[tex]2t=4[/tex]
[tex]t=2[/tex]
[tex](x,y,z) = (1,0,1) + 2(3,-3,2) = (7,-6,5)[/tex]
The intersection is a single point (7,-6,5).
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;
[tex]l = d-a, m = e-b, n = f - c[/tex]
and the equation of the line is given as
[tex]\dfrac{x-a}{l} = \dfrac{y-b}{m} = \dfrac{z-c}{n}[/tex]
Since the two points through which the line passes are (1,0,1) and (4,-3,3)
Thus, we have
[tex]l = 4-1 = 3\\m = -3-0 = -3\\n = 3-1 = 2[/tex]
And thus, equation of line is
[tex]\dfrac{x-4}{3} = \dfrac{y+3}{-3} = \dfrac{z-3}2}[/tex]
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
[tex]\dfrac{x-4}{3} = \dfrac{y+3}{-3} \implies y = 1-x[/tex]
[tex]\dfrac{x-4}{3} = \ \dfrac{z-3}2} \implies z = \dfrac{2x+1}{3}[/tex]
Putting these values in equation of plane, we get:
[tex]x + y + z = 6\\\\x + 1-x + \dfrac{2x+1}{3} = 6\\2x+1 = 15\\x = 7[/tex]
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:
https://brainly.com/question/8889389
