Let $\stackrel{\to <!--\; \to \; -->}{p}=2\stackrel{^<!--\; ^\; -->}{i}+\stackrel{^<!--\; ^\; -->}{j}+3\stackrel{^<!--\; ^\; -->}{k}$ and $\stackrel{\to <!--\; \to \; -->}{q}=\stackrel{^<!--\; ^\; -->}{i}-<!--\; -\; -->\stackrel{^<!--\; ^\; -->}{j}+\stackrel{^<!--\; ^\; -->}{k}$. . If for some real numbers α, β, and λ , we have $15\stackrel{^<!--\; ^\; -->}{i}+10\stackrel{^<!--\; ^\; -->}{j}+6\stackrel{^<!--\; ^\; -->}{k}=\mathrm{\alpha <!--\; \alpha \; -->}(2\stackrel{\to <!--\; \to \; -->}{p}+\stackrel{\to <!--\; \to \; -->}{q})+\mathrm{\beta <!--\; \beta \; -->}(\stackrel{\to <!--\; \to \; -->}{p}-<!--\; -\; -->2\stackrel{\to <!--\; \to \; -->}{q})+\mathrm{\gamma <!--\; \gamma \; -->}(\stackrel{\to <!--\; \to \; -->}{p}\times <!--\; \times \; -->\stackrel{\to <!--\; \to \; -->}{q})$, then the value of γ is ______.

The correct answer is 2.

Let the given vectors be:
$\overrightarrow{p}=2\hat{i}+\hat{j}+3\hat{k}$
and
$\overrightarrow{q}=\hat{i}-\hat{j}+\hat{k}$

We are given the relation:
$15\hat{i}+10\hat{j}+6\hat{k}=\alpha (2\overrightarrow{p}+\overrightarrow{q})+\beta (\overrightarrow{p}-2\overrightarrow{q})+\gamma (\overrightarrow{p}\times \overrightarrow{q})$

Let us represent the left-hand side vector as $\overrightarrow{r}=15\hat{i}+10\hat{j}+6\hat{k}$.

First, we find the cross product $\overrightarrow{p}\times \overrightarrow{q}$:
$\overrightarrow{p}\times \overrightarrow{q}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 1& 3\\ 1& -& 1\end{array}\right|$
$\overrightarrow{p}\times \overrightarrow{q}=\hat{i}(1-(-3\left)\right)-\hat{j}(2-3)+\hat{k}(-2-1)$
$\overrightarrow{p}\times \stackrel{���}{q}=4\hat{i}+\hat{j}-3\hat{k}$

Now, observe that the vector $\overrightarrow{p}\times \overrightarrow{q}$ is perpendicular to both $\overrightarrow{p}$ and $\overrightarrow{q}$. Thus, it is also perpendicular to any linear combination of $\overrightarrow{p}$ and $\overrightarrow{q}$.
Therefore:
$(2\overrightarrow{p}+\overrightarrow{q})\cdot (\overrightarrow{p}\times \overrightarrow{q})=0$
and
$(\overrightarrow{p}-2\overrightarrow{q})\cdot (\overrightarrow{p}\times \overrightarrow{q})=0$

Taking the dot product of both sides of the original vector equation with $\overrightarrow{p}\times \overrightarrow{q}$:
$\overrightarrow{r}\cdot (\overrightarrow{p}\times \overrightarrow{q})=\alpha \cdot 0+\beta \cdot 0+\gamma {|\overrightarrow{p}\times \overrightarrow{q}|}^2$
This simplifies to:
$\gamma =\frac{\overrightarrow{r}\cdot (\overrightarrow{p}\times \overrightarrow{q})}{{|\overrightarrow{p}\times \overrightarrow{q}|}^2}$

Now we calculate the numerator:
$\overrightarrow{r}\cdot (\overrightarrow{p}\times \overrightarrow{q})=(15\hat{i}+10\hat{j}+6\hat{k})\cdot (4\hat{i}+\hat{j}-3\hat{k})$
$\overrightarrow{r}\cdot (\overrightarrow{p}\times \overrightarrow{q})=15\left(4\right)+10\left(1\right)+6(-3)$
$\overrightarrow{r}\cdot (\overrightarrow{p}\times \overrightarrow{q})=60+10-18=52$

Next, we calculate the denominator:
${|\overrightarrow{p}\times \overrightarrow{q}|}^2=4^2+1^2+{(-3)}^2$
${|\overrightarrow{p}\times \overrightarrow{q}|}^2=16+1+9=26$

Substituting these values back into the expression for $\gamma$:
$\gamma =\frac{52}{26}=2$