1. Given points

$A=(3,0,-1),\quad B=(2,10,6),\quad C=(1,2,1)$

We need to find $\cos \angle GOA$, where $O$ is the origin.

So we need the angle between vectors $\overrightarrow{OG}$ and $\overrightarrow{OA}$.

2. Find midpoint $M$ of $AC$

Using midpoint formula:

$M=\left(\frac{3+1}{2},\frac{0+2}{2},\frac{-1+1}{2}\right)=(2,1,0)$

3. Find point $G$ dividing $BM$ in the ratio $2:1$

Since $G$ divides $BM$ in the ratio $2:1$, we interpret this as

$BG:GM=2:1$

Using section formula for internal division of points $B(2,10,6)$ and $M(2,1,0)$:

$G=\left(\frac{2\cdot 2+1\cdot 2}{2+1},\frac{2\cdot 1+1\cdot 10}{2+1},\frac{2\cdot 0+1\cdot 6}{2+1}\right)$

$G=\left(2,4,2\right)$

4. Form vectors $\overrightarrow{OG}$ and $\overrightarrow{OA}$

Since $O=(0,0,0)$,

$\overrightarrow{OG}=(2,4,2), \qquad \overrightarrow{OA}=(3,0,-1)$

5. Use dot product formula

$\cos\theta=\frac{\overrightarrow{OG}\cdot \overrightarrow{OA}}{|\overrightarrow{OG}|\,|\overrightarrow{OA}|}$

First compute dot product:

$\overrightarrow{OG}\cdot \overrightarrow{OA}=2\cdot 3+4\cdot 0+2\cdot(-1)=6-2=4$

Now magnitudes:

$|\overrightarrow{OG}|=\sqrt{2^2+4^2+2^2}=\sqrt{4+16+4}=\sqrt{24}=2\sqrt{6}$

$|\overrightarrow{OA}|=\sqrt{3^2+0^2+(-1)^2}=\sqrt{9+1}=\sqrt{10}$

Thus,

$\cos \angle GOA=\frac{4}{(2\sqrt{6})(\sqrt{10})}$

$=\frac{4}{2\sqrt{60}}=\frac{2}{\sqrt{60}}=\frac{2}{2\sqrt{15}}=\frac{1}{\sqrt{15}}$

6. Compare with options

$\cos \angle GOA=\frac{1}{\sqrt{15}}$

So the correct option is:

A. $\dfrac{1}{\sqrt{15}}$

7. Comparison with stored correct answer

Stored correct answer: A

Our derived answer: A

They match.