Đáp án D

Ta có \[\begin{array}{l}\left| {z + 2w} \right| = 3 \Leftrightarrow {\left| {z + 2w} \right|^2} = 9 \Leftrightarrow \left( {z + 2w} \right).\left( {\overline {z + 2w} } \right) = 9\\ \Leftrightarrow \left( {z + 2w} \right).\left( {\overline z + 2\overline w } \right) = 9 \Leftrightarrow z.\overline z + 2\left( {z.\overline w + \overline z .w} \right) + 4w.\overline z = 9 \Leftrightarrow {\left| z \right|^2} + 2P + 4{\left| w \right|^2} = 9\;\;\;\left( 1 \right)\end{array}\]

Tương tự:

 \[\begin{array}{l}\left| {2z + 3w} \right| = 6 \Leftrightarrow \left( {2z + 3w} \right).\left( {2\overline z + 3\overline w } \right) = 36 \Leftrightarrow 4{\left| z \right|^2} + 6P + 9{\left| w \right|^2} = 36\;\;\;\left( 2 \right)\\\left| {z + 4w} \right| = 7 \Leftrightarrow \left( {z + 4w} \right).\left( {\overline z + 4\overline w } \right) = 49 \Leftrightarrow {\left| z \right|^2} + 4P + 16{\left| w \right|^2} = 49\;\;\;\left( 3 \right)\end{array}\]

Từ (1), (2), (3) ta có \[\left\{ \begin{array}{l}{\left| z \right|^2} = 33\\P = - 28\\{\left| w \right|^2} = 8\end{array} \right. \Rightarrow P = - 28\].