I is not true. Note that the coordinates of $G_1$ and $G_2$ are $\left(\dfrac{-7}{2},2\right)$ and $(1,4)$ respectively. Then, we have
$\begin{array}{rcl}
OG_1 & = & \sqrt{\left(\dfrac{-7}{2}-0\right)^2+(2-0)^2 } \\
OG_1 & = & \dfrac{\sqrt{65}}{2}
\end{array}$
Also,
$\begin{array}{rcl}
OG_2 & = & \sqrt{(1-0)^2+(4-0)^2} \\
OG_2 & = & \sqrt{17}
\end{array}$
Since $OG_1\neq OG_2$, then $\Delta OG_1G_2$ is not a equilateral triangle.
II is true. Sub. $O(0,0)$ into the left side of the equation of $C_2$, we have
$\begin{array}{cl}
& 2(0)^2+2(0)^2-2(0)-16(0)-17 \\
= & -17 \\
< & 0
\end{array}$
Therefore, $O$ lies inside $C_2$.
Sub. $G_1\left(\dfrac{-7}{2},2\right)$ into the left side of the equation of $C_2$, we have
$\begin{array}{cl}
& 2(\dfrac{-7}{2})^2+2(2)^2-2(\dfrac{-7}{2})-16(2)-17 \\
= & \dfrac{-19}{2} \\
< & 0
\end{array}$
Therefore, $G_1$ lies inside $C_2$.
Hence, the line segment $OG_1$ lies inside $C_2$.
III is true. The radius of $C_1$
$\begin{array}{cl}
= & \sqrt{(\dfrac{-7}{2})^2 +(2)^2-15} \\
= & \dfrac{\sqrt{5}}{2}
\end{array}$
The radius of $C_2$
$\begin{array}{cl}
= & \sqrt{1^2+4^2-\dfrac{-17}{2}} \\
= & \sqrt{\dfrac{51}{2}}
\end{array}$
The distance between $G_1$ and $G_2$
$\begin{array}{cl}
= & \sqrt{(\dfrac{-7}{2}-1)^2+(2-4)^2} \\
= & \dfrac{\sqrt{97}}{2}
\end{array}$
Since the distance between $G_1$ and $G_2$ is less than the sum of the radii of $C_1$ and $C_2$, then $C_1$ and $C_2$ intersect at two distinct points.
