Trong không gian Oxyz, cho điểm M(3; -4; -5) và các đường thẳng

Gọi $A\left(-4-5a;4+2a;2+3a\right)\in d_1,B\left(1-b;2+3b;-5-2b\right)\in d_2.$

Vì $M,A,B\in d$ nên chúng thẳng hàng $\Rightarrow \overrightarrow{MA},\overrightarrow{MB}$ cùng phương.

Ta có: $\left\{\begin{array}{l}\overrightarrow{MA}=\left(5a+7;-2a-8;-3a-7\right)\\ \overrightarrow{MB}=\left(b+2;-3b-6;2b\right)\end{array}\right.$

$\Rightarrow \frac{5a+7}{b+2}=\frac{-2a-8}{-3b-6}=\frac{-3a-7}{2b}$

$\iff \left\{\begin{array}{l}15ab+30a+21b+42=2ab+8b+4a+16\\ 10ab+14b=-3ab-6a-7b-14\end{array}\right.$

$\iff \left\{\begin{array}{l}13ab+26a+13b+26=0\\ 13ab+6a+21b+14=0\end{array}\right.$

$\iff \left\{\begin{array}{l}20a-8b+12=0\\ ab+2a+b+2=0\end{array}\right.$

$\iff \left\{\begin{array}{l}5a-2b+3=0\\ ab+2a+b+2=0\end{array}\right.$

$\iff \left\{\begin{array}{l}b=\frac{5a+3}{2}\\ a.\frac{5a+3}{2}+2a+\frac{5a+3}{2}+2=0\end{array}\right.$

$\iff \left\{\begin{array}{l}b=\frac{5a+3}{2}\\ 5a^2+3a+4a+5a+3+4=0\end{array}\right.$

$\iff \left\{\begin{array}{l}b=\frac{5a+3}{2}\\ 5a^2+12a+7=0\end{array}\right.$

$\iff \left[\begin{array}{l}a=-1,b=-1\\ a=-\frac{7}{5},b=-2\end{array}\right.$

$\Rightarrow \left[\begin{array}{l}A\left(1;2;-1\right),B\left(2;-1;-3\right)\\ A\left(3;\frac{6}{5};-\frac{11}{5}\right),B\left(3;-4;-1\right)\end{array}\right.$

TH1: $A\left(1;2;-1\right),B\left(2;-1;-3\right)\Rightarrow \overrightarrow{OA}\left(1;2;-1\right),\overrightarrow{OB}\left(2;-1;-3\right)$

$\Rightarrow S_{OAB}=\frac{1}{2}\left|\left[\overrightarrow{OA},\overrightarrow{OB}\right]\right|=\frac{1}{2}\sqrt{3^2+6^2+0^2}=\frac{3\sqrt{5}}{2}$

TH2: $A\left(3;\frac{6}{5};-\frac{11}{5}\right),B\left(3;-4;-1\right)\Rightarrow \overrightarrow{OA}\left(3;\frac{6}{5};-\frac{11}{5}\right),\overrightarrow{OB}\left(3;-4;-1\right)$

$\Rightarrow S_{OAB}=\frac{1}{2}\left|\left[\overrightarrow{OA},\overrightarrow{OB}\right]\right|=\frac{1}{2}.\sqrt{{10}^2+3,6^2+15,6^2}=\frac{\sqrt{8908}}{10}$.

Chọn B.