Where does the line through the points P(1, 0, 1,) and Q(4, −2, 2) intersect the plane x + y + z = 6?

At P x+y+z=2
At Q x+y+z=4
At 2Q-P (which along with P is the ends of the line segment with a midpoint at Q) which is (2*4-1, 2*-2-0, 2*2-1) which simplifies to (7, -4, 3) we have 7-4+3=6

Same with Nik.

Points on the line are of the form P + k (Q-P) i.e. (1,0,1) + k(3,-2,1), i.e. (1+3k, -2k, 1+k). Plug into the plane and get:

(1+3k) + (-2k) + (1+k) = 6

simplify

k=2

Plug that into (1+3k, -2k, 1+k) and get (7, -4, 3) as the required point.

PQ(3; -2; 1)
(x - 4)/3 = (y+2)/-2 = (z-2)/1
{x = 3t + 4
{y = -2t - 2
{z = t + 2
(3t + 4) + (-2t - 2) + (t + 2) = 6
2t = 2
t = 1
________
x = 7
y = -4
z = 3
Ans. (7; -4; 3)