Given Information
We are given with following vectors:\[{{\bf{v}}_1} = \left[ {\begin{array}{*{20}{r}}1\\{ - 2}\\4\\3\end{array}} \right],{{\bf{v}}_2} = \left[ {\begin{array}{*{20}{r}}4\\{ - 7}\\9\\7\end{array}} \right],{{\bf{v}}_3} = \left[ {\begin{array}{*{20}{r}}5\\{ - 8}\\6\\5\end{array}} \right]\]and \[{\bf{u}} = \left[ {\begin{array}{*{20}{r}}{ - 4}\\{10}\\{ - 7}\\{ - 5}\end{array}} \right]\]We have to determine if the vector $u$ is in the subspace of $R^4$ generated by the set {$v_1, v_2,v_3$}

Step-1:
The vector $u$ will be in the subspace of given set if and only if it satisfies the equation \[{x_1}{v_1} + {x_2}{v_2} + {x_3}{v_3} = u\]

Step-2: The Augmented Matrix
\[M = \left[ {\begin{array}{*{20}{c}}{{v_1}}&{{v_2}}&{{v_3}}&u\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&4&5&{ - 4}\\{ - 2}&{ - 7}&{ - 8}&{10}\\4&9&6&{ - 7}\\3&7&5&{ - 5}\end{array}} \right]\]

Step-3: The Row-reduced Augmented Matrix
\[\begin{array}{l}M = \left[ {\begin{array}{*{20}{c}}1&4&5&{ - 4}\\{ - 2}&{ - 7}&{ - 8}&{10}\\4&9&6&{ - 7}\\3&7&5&{ - 5}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}1&4&5&{ - 4}\\0&1&2&2\\0&{ - 7}&{ - 14}&9\\0&{ - 5}&{ - 10}&7\end{array}} \right]::\,\,\left\{ \begin{array}{l}{R_2} = {R_2} + 2{R_1}\\{R_3} = {R_3} - 4{R_1}\\{R_4} = {R_4} - 3{R_1}\end{array} \right\}\\ = \left[ {\begin{array}{*{20}{c}}1&4&5&{ - 4}\\0&1&2&2\\0&0&0&{23}\\0&0&0&{17}\end{array}} \right]::\,\,\left\{ \begin{array}{l}{R_3} = {R_3} + 7{R_2}\\{R_4} = {R_4} + 5{R_2}\end{array} \right\}\\ = \left[ {\begin{array}{*{20}{c}}1&4&5&{ - 4}\\0&1&2&2\\0&0&0&{23}\\0&0&0&0\end{array}} \right]::\,\,\left\{ {{R_4} = {R_4} - \dfrac{{17}}{{23}}{R_3}} \right\}\end{array}\]We can see that above system does not have any solution, so it is inconsistent. Therefore,

The vector $u$ is not in the subspace of $R^4$ generated by the set {$v_1, v_2,v_3$}