Let $p = 2i + j + 3k$ and $q = i - j + k.$ If for some real numbers α, β and γ, we have $15i + 10j + 6k = α(2p + q) + β(p - 2q) + γ(p × q),$ then the value of γ is ______.

Let p⃗ = 2î + ĵ + 3k̂ and q⃗ = î - ĵ + k̂. If for some real numbers α, β and γ, we have 15î + 10ĵ + 6k̂ = α(2p⃗ + q⃗) + β(p⃗ - 2q⃗) + γ(p⃗ × q⃗), then the value of γ is ________.

$\text{Given: } \vec{p} = 2\hat{i} + \hat{j} + 3\hat{k}, \quad \vec{q} = \hat{i} - \hat{j} + \hat{k}$

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

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

$\text{To find: } \gamma$

$\vec{p}\times\vec{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 3 \\ 1 & -1 & 1 \end{vmatrix}$ $= \hat{i}(1+3) - \hat{j}(2-3) + \hat{k}(-2-1)$ $\vec{p}\times\vec{q} = 4\hat{i} + \hat{j} - 3\hat{k}$ $\vec{p}\cdot(\vec{p}\times\vec{q}) = 0, \quad \vec{q}\cdot(\vec{p}\times\vec{q}) = 0$ $\text{Plan: dot both sides of the equation with } \vec{p}\times\vec{q}$ $(2\vec{p}+\vec{q})\cdot(\vec{p}\times\vec{q}) = 0,\quad (\vec{p}-2\vec{q})\cdot(\vec{p}\times\vec{q}) = 0$ $\vec{v} = 15\hat{i} + 10\hat{j} + 6\hat{k}$ $\vec{v}\cdot(\vec{p}\times\vec{q}) = \alpha\cdot 0 + \beta\cdot 0 + \gamma\,|\vec{p}\times\vec{q}|^{2}$ $\vec{v}\cdot(\vec{p}\times\vec{q}) = (15)(4) + (10)(1) + (6)(-3)$ $= 60 + 10 - 18 = 52$ $|\vec{p}\times\vec{q}|^{2} = 4^{2} + 1^{2} + (-3)^{2} = 16 + 1 + 9 = 26$ $52 = \gamma\,(26) \;\Longrightarrow\; \gamma = \dfrac{52}{26} = 2$ $2\vec{p}+\vec{q} = (4,2,6)+(1,-1,1) = (5,\ 1,\ 7)$ $\vec{p}-2\vec{q} = (2,1,3)-(2,-2,2) = (0,\ 3,\ 1)$ $5\alpha + 4\gamma = 15 \Rightarrow 5\alpha = 7 \Rightarrow \alpha = \dfrac{7}{5}$ $\alpha + 3\beta + \gamma = 10 \Rightarrow 3\beta = \dfrac{33}{5} \Rightarrow \beta = \dfrac{11}{5}$ $7\alpha + \beta - 3\gamma = \dfrac{49}{5} + \dfrac{11}{5} - 6 = 12 - 6 = 6\ \checkmark$ $\gamma = 2$