Let $$\vec{p} = 2\hat{i} + \hat{j} + 3\hat{k}$$ and $$\vec{q} = \hat{i} - \hat{j} + \hat{k}$$. If for some real numbers $$\alpha$$, $$\beta$$ and $$\gamma$$, we have $$15\hat{i} + 10\hat{j} + 6\hat{k} = \alpha(2\vec{p} + \vec{q}) + \beta(\vec{p} - 2\vec{q}) + \gamma(\vec{p} \times \vec{q})$$, then the value of $$\gamma$$ is ______.

Write the given vectors in component form:
$$\vec{p}=2\hat i+\hat j+3\hat k \; \rightarrow \; (2,1,3), \qquad \vec{q}=\hat i-\hat j+\hat k \; \rightarrow \; (1,-1,1).$$

First compute the three auxiliary vectors that appear on the right-hand side.

1. Vector $$2\vec p+\vec q$$
$$2\vec p=(4,2,6) \quad\Longrightarrow\quad 2\vec p+\vec q=(4+1,\,2+(-1),\,6+1)=(5,1,7).$$

2. Vector $$\vec p-2\vec q$$
$$2\vec q=(2,-2,2) \quad\Longrightarrow\quad \vec p-2\vec q=(2-2,\,1-(-2),\,3-2)=(0,3,1).$$

3. Vector $$\vec p\times\vec q$$
Using the determinant rule:
$$ \vec p\times\vec q= \begin{vmatrix} \hat i & \hat j & \hat k\\ 2 & 1 & 3\\ 1 & -1 & 1 \end{vmatrix} = \hat i(1\cdot1-3\cdot(-1))-\hat j(2\cdot1-3\cdot1)+\hat k(2\cdot(-1)-1\cdot1) =4\hat i+\hat j-3\hat k. $$
Thus $$\vec p\times\vec q=(4,1,-3).$$

Denote
$$\vec a_1=(5,1,7),\qquad \vec a_2=(0,3,1),\qquad \vec a_3=(4,1,-3).$$
The equation in component form becomes

$$\alpha\vec a_1+\beta\vec a_2+\gamma\vec a_3=(15,10,6).$$

Equate the $$\hat i, \hat j, \hat k$$ components:

$$5\alpha+4\gamma=15 \quad-(1)$$
$$\alpha+3\beta+\gamma=10 \quad-(2)$$
$$7\alpha+\beta-3\gamma=6 \quad-(3)$$

Solve sequentially. From $$(1):$$
$$\alpha=\frac{15-4\gamma}{5}=3-0.8\gamma.$$

Substitute this $$\alpha$$ into $$(2):$$
$$(3-0.8\gamma)+3\beta+\gamma=10 \;\Longrightarrow\; 3\beta=7-0.2\gamma,$$
so $$\beta=\dfrac{7-0.2\gamma}{3}.$$

Now use $$(3):$$
$$7\alpha+\beta-3\gamma=6.$$ Insert $$\alpha=3-0.8\gamma$$ and the above $$\beta$$:

$$7(3-0.8\gamma)+\beta-3\gamma=6$$ $$\Longrightarrow 21-5.6\gamma+\beta-3\gamma=6$$ $$\Longrightarrow \beta+21-8.6\gamma=6$$ $$\Longrightarrow \beta=-15+8.6\gamma.$$

Equate the two expressions for $$\beta$$:

$$\frac{7-0.2\gamma}{3}=-15+8.6\gamma.$$ Multiply by 3: $$7-0.2\gamma=-45+25.8\gamma$$ $$\Longrightarrow 52=26\gamma$$ $$\Longrightarrow \gamma=2.$$

Hence, the required value is $$\boxed{2}$$.