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

Given
\(\begin{aligned}
& \overline{\mathrm{s}}=4 \overline{\mathrm{p}}+3 \overline{\mathrm{q}}+5 \overline{\mathrm{r}} \\
& \mathrm{~s}=(-\overline{\mathrm{p}}+\overline{\mathrm{q}}+\overline{\mathrm{r}}) x+(\overline{\mathrm{p}}-\overline{\mathrm{q}}+\overline{\mathrm{r}}) y+(-\overline{\mathrm{p}}-\overline{\mathrm{q}}+\overline{\mathrm{r}}) \mathrm{z} \\
& \begin{aligned}
& 4 \overline{\mathrm{p}}+3 \overline{\mathrm{q}}+5 \overline{\mathrm{r}}=(-x+y-\mathrm{z}) \overline{\mathrm{p}}+(x-y-\mathrm{z}) \overline{\mathrm{q}} \\
&+(x+y+\mathrm{z}) \overline{\mathrm{r}}
\end{aligned}
\end{aligned}\)
Comparing, we get
\(\begin{aligned}
\therefore \quad & -x+y-z=4 ...(i)\\
& x-y-z=3 ...(ii)\\
& x+y+z=5
...(iii)\end{aligned}\)
Solving (i), (ii) and (iii), we get
\(x=4, y=\frac{9}{2}, z=\frac{-7}{2}\)
\(\begin{aligned}
\therefore \quad 2 x+y+z & =2(4)+\frac{9}{2}-\frac{7}{2} \\
& =2(4)+1 \\
& =9
\end{aligned}\)