• # question_answer The centroid of a triangle ABC is at the point$\left( \mathbf{1},\mathbf{2},\mathbf{3} \right)$.If the coordinate of A and B are$\left( \mathbf{3},-\mathbf{5},-\mathbf{7} \right)$and$\left( \mathbf{1},\mathbf{7},\mathbf{6} \right)$, respectively then co-ordinate of C. A) $(-1,4,10)$                   B) $\left( 4,-1,10 \right)$     C) $\left( 10,-1,4 \right)$                 D) $\left( 1,4,10 \right)$

[a] Let $C\equiv \left( x,y,z \right)$ According to question,             $1=\frac{3+1+x}{3}\Rightarrow 4+x=3\Rightarrow x=-1$ $\Rightarrow 2=\frac{-5+7+y}{3}$ $\Rightarrow 2+y=6\Rightarrow y=4\And 3=~\frac{-7+6+z}{3}$ $-1+z=9\text{ }z=10.\text{ }C\equiv \left( -1,4,10 \right)$ Hence, option [a]is correct.