Có \(\overrightarrow {AB} = \left( {3;0;3} \right) \Rightarrow AB = 3\sqrt 2 \).

Do \(M \in d \Rightarrow M(5t - 5;4t - 3;t)\)

\( \Rightarrow \left\{ {\begin{array}{*{20}{l}}{\overrightarrow {{\rm{AM}}} = (5{\rm{t}} - 6;4{\rm{t}} - 6;{\rm{t}})}\\{\overrightarrow {{\rm{BM}}} = (5{\rm{t}} - 9;4{\rm{t}} - 6;{\rm{t}} - 3)}\end{array}\quad \Rightarrow \left\{ {\begin{array}{*{20}{l}}{{\rm{M}}{{\rm{A}}^2} = 42{{\rm{t}}^2} - 108{\rm{t}} + 72}\\{{\rm{M}}{{\rm{B}}^2} = 42{{\rm{t}}^2} - 144{\rm{t}} + 126}\\{{\rm{M}}{{\rm{A}}^2} + {\rm{M}}{{\rm{B}}^2} = 84{{\rm{t}}^2} - 252{\rm{t}} + 198\;(*)}\end{array}} \right.} \right.\)

Ta có: \({\rm{A}}{{\rm{B}}^2} = {\rm{M}}{{\rm{A}}^2} + {\rm{M}}{{\rm{B}}^2} - 2{\rm{MA}}.{\rm{MB}}{\rm{.}}\cos \widehat {{\rm{AMB}}}\)

\( \Leftrightarrow 18 = 84{t^2} - 252t + 198 - \sqrt {\left( {42{t^2} - 108t + 72} \right)\left( {42{t^2} - 144t + 126} \right)} \)

\( \Leftrightarrow \sqrt {\left( {42{t^2} - 108t + 72} \right)\left( {42{t^2} - 144t + 126} \right)} = 84{t^2} - 252t + 180\)

\( \Leftrightarrow t = 2\)

Thay \(t = 2\) vào (*), ta được \(T = 30\). Chọn B.