Answer:
(x, y, z) = (1, 1/3, 3 1/3)
Step-by-step explanation:
The normal to plane ABC can be found as the cross product ...
AB×BC = (1, 0, 2)×(2, -2, -3) = (4, 1, 2)
Then the equation of the plane is ...
4x +y +2z = 4·0 +5 +2·3 . . . . using point C to find the constant
4x +y +2z = 11
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The direction vector of the reference line is the vector of coefficients of t: (0, -1, 2). Then the line through point Q is ...
(x, y, z) = (1, 2, 0) +t(0, -1, 2) = (1, 2-t, 2t)
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The value of t that puts a point on this line in plane ABC can be found by substituting these values for x, y, and z in the plane's equation.
4(1) +(2 -t) +2(2t) = 11
Solving for t gives ...
t = 5/3
so the point of intersection of the plane and the line is
(x, y, z) = (1, 2-t, 2t) = (1, 2-5/3, 2·5/3) = (1, 1/3, 3 1/3)
