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 value of $|z_{1} + z_{2} + z_{3}|$ is

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answer is D.

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Detailed Solution

$\begin{array}{l} ∵ |z_{1}| = 1 \Rightarrow {|z_{1}|}^{2} = 1 \\ \Rightarrow z_{1} {\bar{z}}_{1} = 1 \Rightarrow \frac{1}{z_{1}} = {\bar{z}}_{1} \\ |z_{2}| = 2 \Rightarrow {|z_{2}|}^{2} = 4 \Rightarrow z_{2} {\bar{z}}_{2} = 4 \\ \Rightarrow \frac{4}{z_{2}} = {\bar{z}}_{2} and |z_{3}| = 3 \Rightarrow {|z_{3}|}^{2} = 9 \\ \Rightarrow z_{3} {\bar{z}}_{3} = 9 \Rightarrow \frac{9}{z_{3}} = {\bar{z}}_{3} \\ and given |9 z_{1} z_{2} + 4 z_{3} z_{1} + z_{2} z_{3}| = 12 \\ \Rightarrow |z_{1} z_{2} z_{3}| |\frac{9}{z_{3}} + \frac{4}{z_{2}} + \frac{1}{z_{1}}| = 12 \\ \Rightarrow |z_{1}| |z_{2}| |z_{3}| |{\bar{z}}_{3} + {\bar{z}}_{2} + {\bar{z}}_{1}| = 12 \\ [∵ \frac{1}{z_{1}} = {\bar{z}}_{1}, \frac{4}{z_{2}} = {\bar{z}}_{2} and \frac{9}{z_{3}} = {\bar{z}}_{3}] \\ \Rightarrow 1 \cdot 2 \cdot 3 |\bar{z_{1} + z_{2} + z_{3}}| = 12 \\ ∴ |z_{1} + z_{2} + z_{3}| = 2 [∵ | \bar{z} | = | z |] \end{array}$