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$\mathbf{a}, \mathbf{b}, \mathbf{c}$ are non-coplanar vectors. If $\begin{aligned} & \mathbf{a}+3 \mathbf{b}+4 \mathbf{c}=x(\mathbf{a}-2 \mathbf{b}+3 \mathbf{c})+y(\mathbf{a}+5 \mathbf{b}-2 \mathbf{c}) \\ & +z(6 \mathbf{a}+14 \mathbf{b}+4 \mathbf{c}), \text { then } x+y+z=\end{aligned}$
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Verified Answer
The correct answer is:
$-4$
$\mathbf{a}+3 \mathbf{b}+4 \mathbf{c}=\mathbf{a}(x+y+6 z)+\mathbf{b}(-2 x+5 y+14 z)$ $+c(3 x-2 y+4 z)$
(because $a, b$ and $c$ are non-coplanar vectors)
Comparing both sides,
$x+y+6 z=1$ ...(i)
$-2 x+5 y+14 z=3$ ...(ii)
$3 x-2 y+4 z=4$ ...(iii)
By solving Eqs. (i), (ii) and (iii), we get
$\begin{aligned} & x=-2, y=-3, z=1 \\ & x+y+z=-4\end{aligned}$
(because $a, b$ and $c$ are non-coplanar vectors)
Comparing both sides,
$x+y+6 z=1$ ...(i)
$-2 x+5 y+14 z=3$ ...(ii)
$3 x-2 y+4 z=4$ ...(iii)
By solving Eqs. (i), (ii) and (iii), we get
$\begin{aligned} & x=-2, y=-3, z=1 \\ & x+y+z=-4\end{aligned}$
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