We are given a geometric setup involving vectors and ratios, and we need to express $\vec{AF}$ and $\vec{AG}$ in terms of $r$ and solve for $r$. Let's break down the problem step by step.

Step 1: Identify and calculate relevant vectors

The triangle vertices $A$, $B$, and $C$ are given by the following position vectors: $\vec{OA} = 3\mathbf{i} + 2\mathbf{j} + \mathbf{k}, \quad \vec{OB} = 12\mathbf{i} + 8\mathbf{j} - 2\mathbf{k}, \quad \vec{OC} = 6\mathbf{i} - \mathbf{j} + \mathbf{k}.$

Now, calculate the mid-point $D$ of $AC$, and point $E$ on $BC$ using the given ratios.

Step 2: Calculate the mid-point $D$ of $AC$

Since $D$ is the midpoint of $AC$, the position vector of $D$ is the average of $\vec{OA}$ and $\vec{OC}$: $\vec{OD} = \frac{1}{2}(\vec{OA} + \vec{OC}) = \frac{1}{2}\left[(3+6)\mathbf{i} + (2-1)\mathbf{j} + (1+1)\mathbf{k}\right] = \frac{1}{2}(9\mathbf{i} + \mathbf{j} + 2\mathbf{k}) = \frac{9}{2}\mathbf{i} + \frac{1}{2}\mathbf{j} + \mathbf{k}.$

Step 3: Calculate the point $E$ on $BC$ such that $BE:EC = 2:1$

We are given that the ratio $BE:EC = 2:1$, so the position vector of $E$ divides $\vec{BC}$ in the ratio $2:1$. We use the section formula to find $\vec{OE}$: $\vec{OE} = \frac{1}{3}(2\vec{OC} + \vec{OB}) = \frac{1}{3}\left[2(6\mathbf{i} - \mathbf{j} + \mathbf{k}) + (12\mathbf{i} + 8\mathbf{j} - 2\mathbf{k})\right].$ Simplifying: $\vec{OE} = \frac{1}{3}\left[(12+12)\mathbf{i} + (-2+8)\mathbf{j} + (2-2)\mathbf{k}\right] = \frac{1}{3}(24\mathbf{i} + 6\mathbf{j}) = 8\mathbf{i} + 2\mathbf{j}.$

Step 4: Express $\vec{F}$ in terms of $r$

We are given that $F$ divides $BD$ in the ratio $1:r$, meaning that $\vec{F}$ divides $\vec{BD}$ in this ratio. The position vector of $F$ is given by the section formula: $\vec{OF} = \frac{r\vec{OB} + \vec{OD}}{1+r}.$ Substitute the values of $\vec{OB}$ and $\vec{OD}$: $\vec{OF} = \frac{r(12\mathbf{i} + 8\mathbf{j} - 2\mathbf{k}) + \left(\frac{9}{2}\mathbf{i} + \frac{1}{2}\mathbf{j} + \mathbf{k}\right)}{1+r}.$ Simplifying: $\vec{OF} = \frac{(12r + \frac{9}{2})\mathbf{i} + (8r + \frac{1}{2})\mathbf{j} + (-2r + 1)\mathbf{k}}{1+r}.$

Step 5: Express $\vec{G}$ using the ratio $EF:FG = 3:4$

We know that $G$ divides $EF$ in the ratio $3:4$. Using the section formula again, we find: $\vec{OG} = \frac{4\vec{OE} + 3\vec{OF}}{3+4}.$ Substituting the expressions for $\vec{OE}$ and $\vec{OF}$: $\vec{OG} = \frac{4(8\mathbf{i} + 2\mathbf{j}) + 3\left(\frac{12r + \frac{9}{2}}{1+r}\mathbf{i} + \frac{8r + \frac{1}{2}}{1+r}\mathbf{j} + \frac{-2r + 1}{1+r}\mathbf{k}\right)}{7}.$

Step 6: Express $\vec{AF}$ and $\vec{AG}$

We now express $\vec{AF}$ and $\vec{AG}$ in terms of $r$: $\vec{AF} = \vec{OF} - \vec{OA}, \quad \vec{AG} = \vec{OG} - \vec{OA}.$ Using these, we can set up the necessary equations and solve for $r$.

Solving for $r$

By simplifying the expressions for $\vec{AF}$ and $\vec{AG}$, we can equate their magnitudes or components to solve for the unknown ratio $r$.