Let $\mathbf{v}_{1}=\left[\begin{array}{r}-5 \\ 0 \\ 6\end{array}\right], \mathbf{v}_{2}=\left[\begin{array}{r}-5 \\ 3 \\ 3\end{array}\right], \mathbf{v}_{3}=\left[\begin{array}{r}0 \\ -7 \\ 7\end{array}\right]$, and $\mathbf{u}=\left[\begin{array}{r}-7 \\ 7 \\ 3\end{array}\right]$. Determine if $\mathbf{u}$ is in Nul $A$, where $A=\left[\begin{array}{lll}\mathbf{v}_{1} & \mathbf{v}_{2} & \mathbf{v}_{3}\end{array}\right]$

Is $u$ in NulA?
Yes
No

Step 1 ：Form the system of linear equations from the matrix $A$ and vector $\mathbf{u}$: \[\begin{cases} -5x_1 - 5x_2 + 0x_3 = -7 \\ 0x_1 + 3x_2 - 7x_3 = 7 \\ 6x_1 + 3x_2 + 7x_3 = 3 \end{cases}\]

Step 2 ：Form the augmented matrix: \[\left[\begin{array}{ccc|c}-5 & -5 & 0 & -7 \\ 0 & 3 & -7 & 7 \\ 6 & 3 & 7 & 3\end{array}\right]\]

Step 3 ：Divide the first row by -5: \[\left[\begin{array}{ccc|c}1 & 1 & 0 & 1.4 \\ 0 & 3 & -7 & 7 \\ 6 & 3 & 7 & 3\end{array}\right]\]

Step 4 ：Subtract 6 times the first row from the third row: \[\left[\begin{array}{ccc|c}1 & 1 & 0 & 1.4 \\ 0 & 3 & -7 & 7 \\ 0 & -3 & 7 & -5.4\end{array}\right]\]

Step 5 ：Add the second row to the third row: \[\left[\begin{array}{ccc|c}1 & 1 & 0 & 1.4 \\ 0 & 3 & -7 & 7 \\ 0 & 0 & 0 & 1.6\end{array}\right]\]

Step 6 ：The last row of the augmented matrix corresponds to the equation $0=1.6$, which is a contradiction. Therefore, the system of equations has no solution.

Step 7 ：\(\boxed{\text{No, the vector } \mathbf{u} \text{ is not in the null space of matrix } A.}\)