If |z_1|=1, |z_2|=2, |z_3|=3 and |9 z_1 z_2 + | vedclass

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(D) Given that $|z_1|=1, |z_2|=2, |z_3|=3$ and $|9 z_1 z_2 + 4 z_1 z_3 + z_2 z_3| = 12$. We know that $|z|^2 = z \bar{z}$,which implies $\bar{z} = \fr

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(D) Given that $|z_1|=1, |z_2|=2, |z_3|=3$ and $|9 z_1 z_2 + 4 z_1 z_3 + z_2 z_3| = 12$. We know that $|z|^2 = z \bar{z}$,which implies $\bar{z} = \frac{|z|^2}{z}$. Substituting $\bar{z}_1 = \frac{1}{z_1}$,$\bar{z}_2 = \frac{4}{z_2}$,and $\bar{z}_3 = \frac{9}{z_3}$ is not directly helpful here. Instead,divide the given expression by $|z_1 z_2 z_3|$: $|9 z_1 z_2 + 4 z_1 z_3 + z_2 z_3| = 12$ $|z_1 z_2 z_3| \left| \frac{9}{z_3} + \frac{4}{z_2} + \frac{1}{z_1} ight| = 12$ Since $|z_1|=1, |z_2|=2, |z_3|=3$,we have $|z_1 z_2 z_3| = 1 	imes 2 	imes 3 = 6$. $6 \left| \frac{9}{z_3} + \frac{4}{z_2} + \frac{1}{z_1} ight| = 12 \implies \left| \frac{9}{z_3} + \frac{4}{z_2} + \frac{1}{z_1} ight| = 2$. Using $\bar{z} = \frac{|z|^2}{z}$,we have $\bar{z}_1 = \frac{1}{z_1}$,$\bar{z}_2 = \frac{4}{z_2}$,and $\bar{z}_3 = \frac{9}{z_3}$. Thus,$|\bar{z}_1 + \bar{z}_2 + \bar{z}_3| = 2$. Since $|\bar{z}| = |z|$,we get $|z_1 + z_2 + z_3| = 2$.

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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