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If $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are three non-coplanar vectors such that $\mathbf{a}+\mathbf{b}+\mathbf{c}=\alpha \mathbf{d}$ and $\mathbf{b}+\mathbf{c}+\mathbf{d}=\beta \mathbf{a}$, then $\mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d}$ is equal to
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We have $\mathbf{a}+\mathbf{b}+\mathbf{c}=\alpha \mathbf{d}$ and $\mathbf{b}+\mathbf{c}+\mathbf{d}=\beta \mathbf{a}$
$\begin{aligned}
& \therefore \mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d}=(\alpha+1) \mathbf{d} \text { and } \mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d}=(\beta+1) \mathbf{a} . \\
& \Rightarrow(\alpha+1) \mathbf{d}=(\beta+1) \mathbf{a} \\
& \text { If } \alpha \neq-1 \text {, then }(\alpha+1) \mathbf{d}=(\beta+1) \mathbf{a} \Rightarrow \mathbf{d}=\frac{\beta+1}{\alpha+1} \mathbf{a} \\
& \Rightarrow \mathbf{a}+\mathbf{b}+\mathbf{c}=\alpha \mathbf{d} \Rightarrow \mathbf{a}+\mathbf{b}+\mathbf{c}=\alpha\left(\frac{\beta+1}{\alpha+1}\right) \mathbf{a} \\
& \Rightarrow\left(1-\frac{\alpha(\beta+1)}{\alpha+1}\right) \mathbf{a}+\mathbf{b}+\mathbf{c}=0
\end{aligned}$
$\Rightarrow \mathbf{a}, \mathbf{b}, \mathbf{c}$
are coplanar which is contradiction to the given condition,
$\therefore \alpha=-1$ and so $\mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d}=0$.
$\begin{aligned}
& \therefore \mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d}=(\alpha+1) \mathbf{d} \text { and } \mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d}=(\beta+1) \mathbf{a} . \\
& \Rightarrow(\alpha+1) \mathbf{d}=(\beta+1) \mathbf{a} \\
& \text { If } \alpha \neq-1 \text {, then }(\alpha+1) \mathbf{d}=(\beta+1) \mathbf{a} \Rightarrow \mathbf{d}=\frac{\beta+1}{\alpha+1} \mathbf{a} \\
& \Rightarrow \mathbf{a}+\mathbf{b}+\mathbf{c}=\alpha \mathbf{d} \Rightarrow \mathbf{a}+\mathbf{b}+\mathbf{c}=\alpha\left(\frac{\beta+1}{\alpha+1}\right) \mathbf{a} \\
& \Rightarrow\left(1-\frac{\alpha(\beta+1)}{\alpha+1}\right) \mathbf{a}+\mathbf{b}+\mathbf{c}=0
\end{aligned}$
$\Rightarrow \mathbf{a}, \mathbf{b}, \mathbf{c}$
are coplanar which is contradiction to the given condition,
$\therefore \alpha=-1$ and so $\mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d}=0$.
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