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## QuestionMathsClass 12

If the vertices of a triangle are $$(-1,6,-4),(2,1,1)$$ and $$(5,-1,0)$$ then the centroid of the triangle is

(A) $$(6,6,-3)$$
(B) $$(2,2,-1)$$
(C) $$\left ( 3,3,-\displaystyle \frac{3}{2} \right )$$
(D) none of these

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 $$(2,2,-1)$$