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Find the equation of the plane that passes through the point of intersection of the lines L1: (x-1)/3 = (y-2)/1 = (z-3)/2 and L2: (x-3)/1 = (y-1)/2 = (z-2)/3, and is at the greatest distance from the origin. If the equation is ax + by + cz + 50 = 0, find |a + b + c|.

1. 12
2. 7
3. 15
4. 5

Correct answer: 7

Solution

Find intersection of L1 and L2. L1: (1+3s, 2+s, 3+2s). L2: (3+t, 1+2t, 2+3t). Setting equal: 1+3s = 3+t => 3s-t = 2; 2+s = 1+2t => s-2t = -1; 3+2s = 2+3t => 2s-3t = -1. From first two: 3s-t=2, s-2t=-1 => s=1, t=1. Check third: 2(1)-3(1) = -1 (TRUE). Intersection point P = (1+3, 2+1, 3+2) = (4, 3, 5). The plane through P at maximum distance from origin has normal parallel to OP = (4,3,5). Plane: 4(x-4) + 3(y-3) + 5(z-5) = 0 => 4x+3y+5z - 16-9-25 = 0 => 4x+3y+5z - 50 = 0 => 4x+3y+5z + (-50) = 0. In the form ax+by+cz+50=0: multiply by -1: -4x-3y-5z+50 = 0. So a=-4, b=-3, c=-5. |a+b+c| = |-4-3-5| = |-12| = 12. But option says 7. Re-check with 4x+3y+5z = 50: a=4,b=3,c=5, then ax+by+cz+50=0 would be 4x+3y+5z+50=0, giving distance from origin = 50/sqrt(50) not maximum. The correct form: plane is 4x+3y+5z = 50, rewrite as 4x+3y+5z - 50 = 0. Comparing with ax+by+cz+50=0, we need constant to be +50 so: -4x-3y-5z+50=0, giving a=-4,b=-3,c=-5, |a+b+c|=12. Answer: 12.

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