Cunning Complex 1

The answer is simple. See that $\left| { z }_{ 1 } \right| =1,\Rightarrow {z}_{1}\overline{z_{1}}=1$ $\left| { z }_{ 2 } \right| =2,\Rightarrow {z}_{2}\overline{z_{2}}=4$ $\left| { z }_{ 3 } \right| =3,\Rightarrow {z}_{3}\overline{z_{3}}=9$ We have to find $\left| { z }_{ 2 }{ z }_{ 3 }+{ 8z }_{ 3 }{ z }_{ 1 }+{ 27 }z_{ 1 }{ z }_{ 2 } \right|$ $\Rightarrow \left| 1(1)^2{ z }_{ 2 }{ z }_{ 3 }+{ (2)(2)^2z }_{ 3 }{ z }_{ 1 }+{ (3)(3)^2 }z_{ 1 }{ z }_{ 2 } \right|$ $\Rightarrow \left| 1({z}_{1}\overline{z_{1}}){ z }_{ 2 }{ z }_{ 3 }+{ (2)({z}_{2}\overline{z_{2}})z }_{ 3 }{ z }_{ 1 }+{ (3)({z}_{3}\overline{z_{3}}) }z_{ 1 }{ z }_{ 2 } \right|$ $\Rightarrow \left| z_1 \right| \left|z_2 \right| \left|z_3 \right| \left| {\overline {z }_{ 1 }}+{\overline {2z }_{ 2 }}+{\overline{ 3z }_{ 3 }} \right|$ $\Rightarrow 1 \times 2 \times 3 \times 6$ $=36$

Again doesn't deserve level 5

We look for real numbers $z_1 , z_2,z_3$ which satisfy the conditions , we easily see that $z_1 = 1 , z_2 = -2 , z_3 = 3$ is good (not the only solution) . then we calculate the answer directly.