Answer :

If the three points are collinear then the area of the triangle formed by them will be zero.


Area of a Δ ABC whose vertices are A(x1,y1), B(x2,y2) and C(x3,y3) is given by-



Area of given Δ ABC = 0


√(2(k-(-3)) + 4(-3-3) + 6(3-k) ) = 0


squaring both sides, we get-


2(k + 3) + 4(-6) + 6(3-k) = 0


2k + 6-24 + 18-6k = 0


-4k + 24-24 = 0


k = 0


Thus, the value of k is zero.


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