Group Event 7
$O A B C$ is a tetrahedron with $O A, O B$ and $O C$ being mutually perpendicular. Given that $O A=O B=O C=6 x$.
G7.1 If the volume of $O A B C$ is $a x^{3}$, find $a$.
G7.2 If the area of $\triangle A B C$ is $b \sqrt{3} x^{2}$, find $b$.
G7.3 If the distance from $O$ to $\triangle A B C$ is $c \sqrt{3} x$, find $c$.
By finding the volume of $O A B C$ in two different ways.

G7.4 If $\theta$ is the angle of depression from $C$ to the midpoint of $A B$ and $\sin \theta=\frac{\sqrt{d}}{3}$, find $d$.
$$
\begin{array}{l}
a x^{3}=\frac{1}{3} \cdot \frac{1}{2}(6 x)^{2} \cdot(6 x)=36 x^{3} \\
\Rightarrow a=36 \\
A B=B C=A C=\sqrt{(6 x)^{2}+(6 x)^{2}}=6 x \sqrt{2}
\end{array}
$$
$\triangle A B C$ is equilateral
$$
\angle B A C=60^{\circ}
$$

Area of $\triangle A B C=b \sqrt{3} x^{2}=\frac{1}{2}(6 x \sqrt{2})^{2} \sin 60^{\circ}=18 \sqrt{3} x^{2}$
$$
\begin{array}{l}
b=18 \\
\frac{1}{3} \cdot 18 \sqrt{3} x^{2} \times(c \sqrt{3} x)=36 x^{3} \\
c=2 \\
\frac{1}{3} \cdot 18 \sqrt{3} x^{2} \times(c \sqrt{3} x)=36 x^{3}
\end{array}
$$

Let the midpoint of $A B$ be $M$.
$$
\begin{array}{l}
O C=6 x, \frac{O M \times A B}{2}=\frac{O A \times O B}{2} \\
\Rightarrow 6 x \sqrt{2} \cdot O M=(6 x)^{2} \\
\Rightarrow O M=3 \sqrt{2} x \\
C M=\sqrt{O M^{2}+O C^{2}} \\
=\sqrt{(3 \sqrt{2} x)^{2}+(6 x)^{2}} \\
=3 \sqrt{6} x \\
\sin \theta=\frac{\sqrt{d}}{3}=\frac{O C}{C M} \\
=\frac{6 x}{3 \sqrt{6} x}=\frac{\sqrt{6}}{3} \\
d=6 \\
\end{array}
$$