a)Do $M\in
Ox\Rightarrow M(a;0;0)\Rightarrow MA=\sqrt{(a-1)^2+4+9}=\sqrt{a^2-2a+14}$
$MB=\sqrt{(a+3)^2+9+4}=\sqrt{a^2+6a+22}$
Để M cách đều
A, B$\Leftrightarrow MA=MB\Leftrightarrow a^2-2a+14=a^2+6a+22\Leftrightarrow
a=-1$
$M(-1;0;0)$.
b) Để $AB$ vuông góc với $OC$ điều kiện là:
$\overrightarrow {AB} \bot \overrightarrow {OC} \Leftrightarrow \overrightarrow {AB} \bot \overrightarrow {OC}=0 \Leftrightarrow (2;\sqrt{3};1).(\sin 5t; \cos 3t; \sin 3t)=0$
$\Leftrightarrow 2\sin 5t+ \sqrt{3}\cos 3t+\sin 3t=0 \Leftrightarrow \sin 5t=-\frac{\sqrt{3}}{2}.\cos 3t- \frac{1}{2}.\sin 3t$
$\Leftrightarrow \sin 5t=-(\sin \frac{\pi}{3}.\cos 3t+\cos \frac{\pi}{3}.\sin 3t)=-\sin (3t+\frac{\pi}{3})=\sin (-3t-\frac{\pi}{3})$
$\Leftrightarrow \left[ \begin{array}{l} 5t=-3t-\frac{\pi}{3}+2k\pi\\
5t=\pi+3t+\frac{\pi}{3}+2k\pi \end{array} \right. \Leftrightarrow
\left[ \begin{array}{l} t=-\frac{\pi}{24}+\frac{k\pi}{4}\\
t=\frac{2\pi}{3}+k\pi \end{array} \right. ,k\in Z $