Biến đổi phương trình tương đương: \(i({\rm{a}} + {\rm{bi}} - 5) = 2({\rm{a}} - {\rm{bi}} - 3) - (1 - {\rm{i}})\sqrt {{{\rm{a}}^2} + {{\rm{b}}^2}} \)

\( \Leftrightarrow \left( {2{\rm{a}} + {\rm{b}} - 6 - \sqrt {{{\rm{a}}^2} + {{\rm{b}}^2}} } \right) + \left( {\sqrt {{{\rm{a}}^2} + {{\rm{b}}^2}} - {\rm{a}} - 2\;{\rm{b}} + 5} \right){\rm{i}} = 0\)

\( \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{2{\rm{a}} + {\rm{b}} - 6 - \sqrt {{{\rm{a}}^2} + {{\rm{b}}^2}} = 0}\\{\sqrt {{{\rm{a}}^2} + {{\rm{b}}^2}} - {\rm{a}} - 2{\rm{b}} + 5 = 0}\end{array} \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{\sqrt {{{\rm{a}}^2} + {{\rm{b}}^2}} = 2{\rm{a}} + {\rm{b}} - 6}\\{\sqrt {{{\rm{a}}^2} + {{\rm{b}}^2}} = {\rm{a}} + 2\;{\rm{b}} - 5}\end{array} \Rightarrow 2{\rm{a}} + {\rm{b}} - 6 = {\rm{a}} + 2\;{\rm{b}} - 5} \right.} \right. \Leftrightarrow b = a - 1\).

Khi đó ta có: \(\sqrt {{a^2} + {{(a - 1)}^2}} = 2a + (a - 1) - 6 \Leftrightarrow \sqrt {2{a^2} - 2a + 1} = 3a - 7\)

\( \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{a \ge \frac{7}{3}}\\{7{a^2} - 40a + 48 = 0}\end{array}\quad \Leftrightarrow a = 4 \Rightarrow b = 3 \Rightarrow T = a - 2b = - 2} \right.\). Chọn C.