Let $\vec{a}, \vec{b}, \vec{c}$ be three vectors in the xyz space such that $\vec{a} \times \vec{b}=\vec{b} \times \vec{c}=\vec{c} \times \vec{a} \neq 0$ If $A, B, C$ are points with position vectors $\vec{a}, \vec{b}, \vec{c}$ respectively, then the number of possible positions of the centroid of triangle $A B C$ is -

• Correct answer
  A

  1
• B

  2
• C

  3
• D

  6

Solution

$\begin{array}{lr}\vec{a} \times \vec{b}+\vec{c} \times \vec{b}=0 & \text { similarly } \quad \vec{b}+\vec{c}=\lambda_{2} \vec{a} \\ \vec{a}+\vec{c}=\lambda_{1} \vec{b} & \vec{b}+\vec{a}=\lambda_{3} \vec{c}\end{array}$

Hence, the correct answer is option: 'A'.