Let vec{p}=2 hat{i}+hat{j}+3 hat{k} and vec{q | vedclass

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(B) Given the equation: $15 \hat{i}+10 \hat{j}+6 \hat{k}=\alpha(2 \vec{p}+\vec{q})+\beta(\vec{p}-2 \vec{q})+\gamma(\vec{p} 	imes \vec{q})$.Taking the

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(B) Given the equation: $15 \hat{i}+10 \hat{j}+6 \hat{k}=\alpha(2 \vec{p}+\vec{q})+\beta(\vec{p}-2 \vec{q})+\gamma(\vec{p} 	imes \vec{q})$.Taking the dot product of both sides with $(\vec{p} 	imes \vec{q})$,we note that $(\vec{p} 	imes \vec{q}) \cdot (2 \vec{p}+\vec{q}) = 0$ and $(\vec{p} 	imes \vec{q}) \cdot (\vec{p}-2 \vec{q}) = 0$ because the cross product is perpendicular to both vectors.Thus,we have: $(15 \hat{i}+10 \hat{j}+6 \hat{k}) \cdot (\vec{p} 	imes \vec{q}) = \gamma |\vec{p} 	imes \vec{q}|^2$.The left side is the scalar triple product $[15 \hat{i}+10 \hat{j}+6 \hat{k}, \vec{p}, \vec{q}]$,which is given by the determinant:$\begin{vmatrix} 15 & 10 & 6 \ 2 & 1 & 3 \ 1 & -1 & 1 \end{vmatrix} = 15(1 - (-3)) - 10(2 - 3) + 6(-2 - 1) = 15(4) - 10(-1) + 6(-3) = 60 + 10 - 18 = 52$.Now,calculate $\vec{p} 	imes \vec{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 1 & 3 \ 1 & -1 & 1 \end{vmatrix} = \hat{i}(1 - (-3)) - \hat{j}(2 - 3) + \hat{k}(-2 - 1) = 4 \hat{i} + \hat{j} - 3 \hat{k}$.Then $|\vec{p} 	imes \vec{q}|^2 = 4^2 + 1^2 + (-3)^2 = 16 + 1 + 9 = 26$.Substituting these into the equation: $52 = \gamma(26)$.Therefore,$\gamma = 2$.

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.

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