Suppose that $\vec{p}, \vec{q}$ and $\vec{r}$ are three non-coplanar vectors in $\mathbb{R}^{3}$. Let the components of a vector $\vec{s}$ along $\vec{p}, \vec{q}$ and $\vec{r}$ be $4$, $3$ and $5$, respectively. If the components of this vector $\vec{s}$ along $(-\vec{p}+\vec{q}+\vec{r}), (\vec{p}-\vec{q}+\vec{r})$ and $(-\vec{p}-\vec{q}+\vec{r})$ are $x$, $y$ and $z$, respectively, then the value of $2x+y+z$ is

$\text{Given: } \vec{p}, \vec{q}, \vec{r} \text{ non-coplanar in } \mathbb{R}^3$ $\text{components of } \vec{s} \text{ along } \vec{p}, \vec{q}, \vec{r} \text{ are } 4,\ 3,\ 5$ $\text{To find: } 2x + y + z$ $x,\ y,\ z = \text{components of } \vec{s} \text{ along } (-\vec{p}+\vec{q}+\vec{r}),\ (\vec{p}-\vec{q}+\vec{r}),\ (-\vec{p}-\vec{q}+\vec{r})$ $\vec{p},\ \vec{q},\ \vec{r}\ \text{non-coplanar} \Rightarrow \text{basis of } \mathbb{R}^3$ $\vec{s}=4\vec{p}+3\vec{q}+5\vec{r}$ $\vec{s}=x\vec{a}+y\vec{b}+z\vec{c}$ $\text{Plan: expand in } \vec{p},\vec{q},\vec{r} \text{ and compare coefficients}$ $\vec{a}=-\vec{p}+\vec{q}+\vec{r}$ $\vec{b}=\vec{p}-\vec{q}+\vec{r}$ $\vec{c}=-\vec{p}-\vec{q}+\vec{r}$ $x\vec{a}+y\vec{b}+z\vec{c}=x(-\vec{p}+\vec{q}+\vec{r})+y(\vec{p}-\vec{q}+\vec{r})+z(-\vec{p}-\vec{q}+\vec{r})$ $=(-x+y-z)\vec{p}+(x-y-z)\vec{q}+(x+y+z)\vec{r}$ $-x+y-z=4$ $x-y-z=3$ $x+y+z=5$ $(-x+y-z)+(x-y-z)=4+3$ $-2z=7 \Rightarrow z=-\dfrac{7}{2}$ $(x-y-z)+(x+y+z)=3+5$ $2x=8 \Rightarrow x=4$ $x+y+z=5 \Rightarrow 4+y-\dfrac{7}{2}=5$ $y=\dfrac{9}{2}$ $2x+y+z=2(4)+\dfrac{9}{2}-\dfrac{7}{2}$ $=8+1=9$ $-x+y-z=-4+\dfrac{9}{2}+\dfrac{7}{2}=-4+8=4 \ \checkmark$ $\boxed{2x+y+z=9}$