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If $\bar{a}, \bar{b}, \bar{c}$ are the non-coplanar vectors and $\bar{a}-2 \bar{b}+3 \bar{c}$, $-4 \bar{a}+5 \bar{b}-6 \bar{c}, x \bar{a}-9 \bar{b}+z \bar{c}$ are collinear points then $2 x-z=$
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Let points $\mathrm{A}_{-}^{\mathrm{A}}:(\overline{\mathrm{a}}-2 \overline{\mathrm{b}}+3 \overline{\mathrm{c}}), \mathrm{B}:(-4 \overline{\mathrm{a}}+5 \overline{\mathrm{b}}-6 \overline{\mathrm{c}})$ and $C:(x \bar{a}-9 \bar{b}+z \bar{c})$
Since, points $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are collinear
$\begin{aligned}
& \Rightarrow \overline{\mathrm{AB}}=\lambda \overline{\mathrm{AC}} \\
& \Rightarrow-5 \overline{\mathrm{a}}+7 \overline{\mathrm{b}}-9 \overline{\mathrm{c}}=\lambda((\mathrm{x}-1) \overline{\mathrm{a}}-7 \overline{\mathrm{b}}+(\mathrm{z}-3) \overline{\mathrm{c}})
\end{aligned}$
On comparing we get, $\lambda=-1, x=6$ and $z=12$
Hence, $2 \mathrm{x}-\mathrm{z}=0$
Since, points $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are collinear
$\begin{aligned}
& \Rightarrow \overline{\mathrm{AB}}=\lambda \overline{\mathrm{AC}} \\
& \Rightarrow-5 \overline{\mathrm{a}}+7 \overline{\mathrm{b}}-9 \overline{\mathrm{c}}=\lambda((\mathrm{x}-1) \overline{\mathrm{a}}-7 \overline{\mathrm{b}}+(\mathrm{z}-3) \overline{\mathrm{c}})
\end{aligned}$
On comparing we get, $\lambda=-1, x=6$ and $z=12$
Hence, $2 \mathrm{x}-\mathrm{z}=0$
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