The point A(1, - 1, 3), B(2, - 4,8) and C(5,- 18, 11) are

Now,Δxy=121−112−415−131=12|[1(−4+13)+1(2−5)+1(−26+20)]|=12|[9−3−6]|=0Δyz=12−131−451−13111=12|[−1(5−11)−3(−4+13)+1(−44+65)]|=12|[6−27+21]|=0Δzx=123115211151=12|[3(2−5)−1(5−11)+1(25−22)]|=12|[−9+6+3]|=0∴ Area of triangle=Δxy2+Δyz2+Δzx2=0+0+0=0Hence, given points are collinear.