The given points A(k,3),B(6,−2), and C(−3,4) are collinear if area of the △ABC formed by joining these points will be zero.
or, area △ABC=0
⇒12|[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]|=0
⇒12|[k(−2−4)+6(4−3)+(−3)(3+2)]|=0
⇒12|[k(−6)+6−15]|=0
⇒|[−6k−9]|=0
⇒k=9−6
⇒k=−32