Question

If $\left|z_{1}\right|=1,\left|z_{2}\right|=2,\left|z_{3}\right|=3$ and $\left|9 z_{1} z_{2}+4 z_{1} z_{3}+z_{2} z_{3}\right|=12$, then the value of $\left|z_{1}+z_{2}+z_{3}\right|$ is

• A. 3
• B. 4
• C. 8
• D. 2

Answer 1

Answer: (D)

Given, $\left|z_{1}\right|=1,\left|z_{2}\right|=2,\left|z_{3}\right|=3$ and
$\left|9 z_{1} z_{2}+4 z_{1} z_{3}+z_{2} z_{3}\right|=12$
Now, $\left|9 z_{1} z_{2}+4 z_{1} z_{3}+z_{2} z_{3}\right|=12$
$\Rightarrow |z_{3}|^{2} z_{1} z_{2}+|z_{2}|^{2} z_{1} z_{3} +|z_{1}|^{2} z_{2} z_{3}|=12$
$\Rightarrow z_{3} \bar{z}_{3} z_{1} z_{2}+z_{2} \bar{z}_{2} z_{1} z_{3} +z_{1} \bar{z}_{1} z_{2} z_{3}=12 \,\,\,\left[\because|z|^{2}=z \bar{z}\right]$
$\Rightarrow \left|z_{1} z_{2} z_{3}\left(\bar{z}_{3}+\bar{z}_{2}+\bar{z}_{1}\right)\right|=12$
$\Rightarrow \left|z_{1} z_{2} z_{3}\right|\left|\bar{z}_{1}+\bar{z}_{2}+\bar{z}_{3}\right|=12$
$\Rightarrow \left|z_{1}\right|\left|z_{2}\right|\left|z_{3}\right|\left|\bar{z}_{1}+\bar{z}_{2}+\bar{z}_{3}\right|=12$
$\Rightarrow 1 \times 2 \times 3\left|\bar{z}_{1}+\bar{z}_{2}+\bar{z}_{3}\right|=12$
$\Rightarrow \left|\bar{z}_{1}+\bar{z}_{2}+\bar{z}_{3}\right|=2$
$\therefore \left|z_{1}+z_{2}+z_{3}\right|=2$