Let $Z_{1}, Z_{2}$ and $Z_{3}$ represents vertices of any triangle ABC, respectively. If $|Z_{1}| = 1, |Z_{2}| = 2, |Z_{3}| = 3$ and $|9 Z_{1} Z_{2} + 4 Z_{1} Z_{3} + Z_{2} Z_{3}| = 12$ then the distance of origin from the centroid of $ΔABC$ is _____

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$\frac{4}{3}$

$\frac{2}{3}$

answer is C.

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Detailed Solution

$\begin{array}{l} {|z_{1}|}^{2} = 1 \Rightarrow z_{1} {\bar{z}}_{1} = 1, {|z_{2}|}^{2} = 4 \Rightarrow z_{2} {\bar{z}}_{2} = 4, z_{3} {\bar{z}}_{3} = 9 \\ |9 z_{1} z_{2} + 4 z_{1} z_{3} + z_{2} z_{3}| = 12 \\ |z_{3} {\bar{z}}_{3} z_{1} z_{2} + z_{2} {\bar{z}}_{2} z_{1} z_{3} + {\bar{z}}_{1} z_{1} z_{2} z_{3}| = 12 \\ |z_{1} z_{2} z_{3} ({\bar{z}}_{1} + {\bar{z}}_{2} + {\bar{z}}_{3})| = 12 \\ |z_{1}| |z_{2}| |z_{3}| |z_{1} + z_{2} + z_{3}| = 12 \\ (1) (2) (3) |z_{1} + z_{2} + z_{3}| = 12 \\ = |z_{1} + z_{2} + z_{3}| = 2 \end{array}$