Given Information
We are given with following vectors \[{v_1} = \left[ {\begin{array}{*{20}{c}}{ - 3}\\0\\6\end{array}} \right],{v_2} = \left[ {\begin{array}{*{20}{c}}{ - 2}\\2\\3\end{array}} \right],{v_3} = \left[ {\begin{array}{*{20}{c}}0\\{ - 6}\\3\end{array}} \right],{\rm{ }}\,\,{\rm{and }}\,\,p = \left[ {\begin{array}{*{20}{c}}1\\{14}\\{ - 9}\end{array}} \right]\]we have to find if the vector $p$ is in Column space of A formed by vectors $v_1, v_2, v_3$ . We will find the augmented matrix formed by the matrix A and the vector $p$. By row-reducing the matrix, we can find whether the vector $p$ is in columns space of A.

Step-1: The Augmented matrix
\[[A\,\,\,p] = \left[ {\begin{array}{*{20}{c}}{ - 3}&{ - 2}&0&1\\0&2&{ - 6}&{14}\\6&3&3&{ - 9}\end{array}} \right]\]

Step- 2: Row-Reduced Form of Augmented matrix
\[\begin{array}{l}[A\,\,\,p] = \left[ {\begin{array}{*{20}{c}}{ - 3}&{ - 2}&0&1\\0&2&{ - 6}&{14}\\6&3&3&{ - 9}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{ - 3}&{ - 2}&0&1\\0&2&{ - 6}&{14}\\0&{ - 1}&3&{ - 7}\end{array}} \right]::\,\,\left\{ {{R_3} \to {R_3} + 2{R_1}} \right\}\\ = \left[ {\begin{array}{*{20}{c}}{ - 3}&{ - 2}&0&1\\0&2&{ - 6}&{14}\\0&0&0&0\end{array}} \right]::\,\,\left\{ {{R_3} \to {R_3} + \dfrac{1}{2}{R_2}} \right\}\end{array}\]As the last row has all elements equal to zero,

The vector $p$ is in Columns space of A.