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Determine the value of k such that the points (4,-2,6), B(0,1,0) and C(1,0,-5) and D (1,k,-2) lie on the same plane.

(B-A)x(C-A) = {-4,3,-6}x{-3,2,-11} = {-21,-26,1}

If D is in the plane of ABC, then (D-A)x(C-A) will also be normal to the plane of ABC. That is,

{-3,k+2,-8}x{-3,2,-11} = {-21,-26,1}
The plane is
21x+26y-z = 26

So,

21+26k+2 = 26
26k = 3
k = 3/26

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