Answer: B
the centroid of the triangle is\((\dfrac {x_1+x_2+x_3}{3},\dfrac{y_1+y_2+y_3}{3},\dfrac{z_1+z_2+z_3}{3})\)
Thus by substituting the vertices we get
\(=(\dfrac{-1+2+5}{3},\dfrac{6+1-1}{3},\dfrac{-4+1+0}{3})\)
\(=(\dfrac{6}{3},\dfrac{6}{3},\dfrac{-3}{3})\)
\(\therefore\)the centroid of the triangle is \((2,2,-1)\)