Solution:
a - 2b + 3c
$\frac{\text{3a}+4\text{b}-5\text{c}}{3}$
$\overrightarrow{\text{c}}=\frac{2\overrightarrow{\text{b}}+\overrightarrow{\text{a}}}{3}$
$\overrightarrow{\text{b}}=\frac{3\overrightarrow{\text{c}}-\overrightarrow{\text{a}}}{2}$
$=\frac{\big(3\overrightarrow{a}+4\overrightarrow{b}-5\overrightarrow{c}\big)-\big(\overrightarrow{a}+2\overrightarrow{b}-3\overrightarrow{c}\big)}{2}$
$=\overrightarrow{a}+3\overrightarrow{b}-4\overrightarrow{c}$
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If sets A and B are defined as $\text{A}=\Big\{(\text{x},\text{y})|\text{y}=\frac{1}{\text{x}},0\neq\text{x}\in\text{R}\Big\}\ \text{B}=\{(\text{x},\text{y})|\text{y}=-\text{x},\text{x}\in\text{R}\},$ then
$\text{A}\cap\text{B}=\text{A}$
$\text{A}\cap\text{B}=\text{B}$
$\text{A}\cap\text{B}=\phi$
$\text{A}\cup\text{B}=\text{A}$