Suppose that vec{p}, vec{q} and vec{r} are th | vedclass

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(D) Given that $\vec{s} = 4\vec{p} + 3\vec{q} + 5\vec{r}$.Also,$\vec{s} = x(-\vec{p} + \vec{q} + \vec{r}) + y(\vec{p} - \vec{q} + \vec{r}) + z(-\vec{p

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$9$

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(D) Given that $\vec{s} = 4\vec{p} + 3\vec{q} + 5\vec{r}$.Also,$\vec{s} = x(-\vec{p} + \vec{q} + \vec{r}) + y(\vec{p} - \vec{q} + \vec{r}) + z(-\vec{p} - \vec{q} + \vec{r})$.Expanding the right side,we get $\vec{s} = (-x + y - z)\vec{p} + (x - y - z)\vec{q} + (x + y + z)\vec{r}$.Since $\vec{p}, \vec{q}, \vec{r}$ are non-coplanar,they are linearly independent. Comparing coefficients:$-x + y - z = 4$  $(1)$$x - y - z = 3$  $(2)$$x + y + z = 5$  $(3)$Adding $(2)$ and $(3)$,we get $2x = 8 \Rightarrow x = 4$.Substituting $x = 4$ in $(3)$,$y + z = 1$.Substituting $x = 4$ in $(2)$,$4 - y - z = 3 \Rightarrow y + z = 1$ (consistent).Substituting $x = 4$ in $(1)$,$-4 + y - z = 4 \Rightarrow y - z = 8$.Solving $y + z = 1$ and $y - z = 8$,we get $2y = 9 \Rightarrow y = 4.5$ and $2z = -7 \Rightarrow z = -3.5$.Finally,$2x + y + z = 2(4) + 4.5 - 3.5 = 8 + 1 = 9$.

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