$\text{Let }\overrightarrow{p}=2\hat{i}+\hat{j}+3\widehat{\mathrm{k }}\text{and}\overrightarrow{\mathrm{ q}}=\hat{i}-\hat{j}+\hat{k}.$ $\text{If for some real numbers}\mathrm{ \alpha },\beta ,\text{and}\gamma ,\text{we have}$

$15\hat{i}+10\hat{j}+6\hat{k}=\alpha (2\overrightarrow{p}+\overrightarrow{q})+\beta (\overrightarrow{p}-2\overrightarrow{q})+\gamma (\overrightarrow{p}\times \overrightarrow{q}),$

$\text{then the value of }\gamma \text{ is \_\_\_\_\_}$

The correct answer is 2.

Step 1: Understand the given vectors and equation
We are given two vectors:

$\overrightarrow{p}=2\hat{i}+\hat{j}+3\hat{k}$

$\overrightarrow{q}=\hat{i}-\hat{j}+\hat{k}$

Let $\overrightarrow{r}$ be the given resultant vector:

$\overrightarrow{r}=15\hat{i}+10\hat{j}+6\hat{k}$

We are given the vector equation:

$\overrightarrow{r}=\alpha (2\overrightarrow{p}+\overrightarrow{q})+\beta (\overrightarrow{p}-2\overrightarrow{q})+\gamma (\overrightarrow{p}\times \overrightarrow{q})$

Step 2: Calculate the cross product $\overrightarrow{p}\times \overrightarrow{q}$
Using the determinant method for vector cross product:

$\overrightarrow{p}\times \overrightarrow{q}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 1& 3\\ 1& \mathrm{-1}& 1\end{array}\right|$

$\overrightarrow{p}\times \overrightarrow{q}=\hat{i}(1-(\mathrm{-3}\left)\right)-\hat{j}(2-3)+\hat{k}(\mathrm{-2}-1)$

$\overrightarrow{p}\times \overrightarrow{q}=4\hat{i}+\hat{j}-3\hat{k}$

Step 3: Simplify using vector orthogonality
Notice that both vectors $(2\overrightarrow{p}+\overrightarrow{q})$ and $(\overrightarrow{p}-2\overrightarrow{q})$ lie in the plane containing $\overrightarrow{p}$ and $\overrightarrow{q}$.
Since the cross product $\overrightarrow{p}\times \overrightarrow{q}$ is perpendicular to every vector in the plane containing $\overrightarrow{p}$ and $\overrightarrow{q}$, their dot products with $\overrightarrow{p}\times \overrightarrow{q}$ are zero:

$(2\overrightarrow{p}+\overrightarrow{q})·(\overrightarrow{p}\times \overrightarrow{q})=0$

$(\overrightarrow{p}-2\overrightarrow{q})·(\overrightarrow{p}\times \overrightarrow{q})=0$

Taking the dot product of both sides of the main equation with $(\overrightarrow{p}\times \overrightarrow{q})$:

$\overrightarrow{r}·(\overrightarrow{p}\times \overrightarrow{q})=\gamma {|\overrightarrow{p}\times \overrightarrow{q}|}^2$

Step 4: Compute the terms and solve for $\gamma$
Calculate $\overrightarrow{r}·(\overrightarrow{p}\times \overrightarrow{q})$:

$\overrightarrow{r}·(\overrightarrow{p}\times \overrightarrow{q})=\left(15\right)\left(4\right)+\left(10\right)\left(1\right)+\left(6\right)\left(\mathrm{-3}\right)$

$\overrightarrow{r}·(\overrightarrow{p}\times \overrightarrow{q})=60+10-18=52$

Calculate ${|\overrightarrow{p}\times \overrightarrow{q}|}^2$:

${|\overrightarrow{p}\times \overrightarrow{q}|}^2=4^2+1^2+{\left(\mathrm{-3}\right)}^2=16+1+9=26$

Substitute these back into the relation:

$52=\gamma \times 26$

$\gamma =\frac{52}{26}=2$

Thus, the value of $\gamma$ is 2.