A line through (1,-1,2) with direction ratios \langle 3, 2, 2 \rangle meets the plane x+2y+3z=18. What is the point of intersection of line and plane?

Explanation

The equation of the line is \frac{x-1}{3} = \frac{y+1}{2} = \frac{z-2}{2} = k. Any generic point on this line is (3k+1, 2k-1, 2k+2). Substitute this point into the plane equation x+2y+3z=18: (3k+1) + 2(2k-1) + 3(2k+2) = 18 \implies 3k+1+4k-2+6k+6 = 18 \implies 13k+5 = 18 \implies 13k = 13 \implies k=1. The point of intersection is (3(1)+1, 2(1)-1, 2(1)+2) = (4, 1, 4).

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