Step 1
Given Matrix, A and vector, $\bf{b}$:
\[\begin{array}{l}A = \left[ {\begin{array}{*{20}{c}}1&{ - 4}&2\\0&3&5\\{ - 2}&8&{ - 4}\end{array}} \right]\\\\{\bf{b}} = \left[ {\begin{array}{*{20}{c}}3\\{ - 7}\\{ - 3}\end{array}} \right]\end{array}\]We have to determine, whether the vector $\bf{b}$ is a linear combination of vectors formed from the columns of the matrix A.

Step 2: Form the vectors from the columns of matrix A
\[\begin{array}{l}{{\bf{a}}_{\bf{1}}} = \left[ {\begin{array}{*{20}{c}}1\\0\\{ - 2}\end{array}} \right]\\\\{{\bf{a}}_2} = \left[ {\begin{array}{*{20}{c}}{ - 4}\\3\\8\end{array}} \right]\\\\{{\bf{a}}_3} = \left[ {\begin{array}{*{20}{c}}2\\5\\{ - 4}\end{array}} \right]\end{array}\]

Step 3: Augmented Form
Convert the vectors ($\bf{a_1, a_2, a_3}$ and $\bf{b}$) into augmented form.\[\begin{array}{l}\left[ {\begin{array}{*{20}{c}}{{{\bf{a}}_{\bf{1}}}}&{{{\bf{a}}_{\bf{2}}}}&{{{\bf{a}}_{\bf{3}}}}&{\bf{b}}\end{array}} \right]\\\\\left[ {\begin{array}{*{20}{c}}1&{ - 4}&2&3\\0&3&5&{ - 7}\\{ - 2}&8&{ - 4}&{ - 3}\end{array}} \right]\end{array}\]

Step 4: Convert to echelon form
\[\begin{array}{l}\left[ {\begin{array}{*{20}{c}}1&{ - 4}&2&3\\0&3&5&{ - 7}\\{ - 2}&8&{ - 4}&{ - 3}\end{array}} \right]\\\\~\left[ {\begin{array}{*{20}{c}}1&{ - 4}&2&3\\0&3&5&{ - 7}\\{ - 2 + 2\left( 1 \right)}&{8 + 2\left( { - 4} \right)}&{ - 4 + 2\left( 2 \right)}&{ - 3 + 2\left( { - 3} \right)}\end{array}} \right]\\\\~\left[ {\begin{array}{*{20}{c}}1&{ - 4}&2&3\\0&3&5&{ - 7}\\0&0&0&3\end{array}} \right]\end{array}\]

Step 5: Equivalent system of Equations
\[\begin{array}{l}{x_1} - 4{x_2} + 2{x_3} = 3\\\,\,\,\,\,\,\,\,\,\,3{x_2} + 5{x_3} = - 7\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 = 3\end{array}\]

Step 6: ANSWER
From the last, row we can see that the system has no solution, as $0 \ne 3$,
The vector $\bf{b}$ is not a linear combination of vectors formed by columns of matrix A