I have found the rank of M, the basis for the null space and evaluated M$\begin{pmatrix} 1\\ -2\\ -3\\ -4\end{pmatrix}$.
But, I am having some trouble answering the last part of the question. Could someone please guide me on how to show that every solution of Mx $=\begin{pmatrix} 2\\ 16\\ 10\\ 22\end{pmatrix}$ has the form as shown in the question? I tried to find something to read on google but I am not sure which section of linear algebra is this on.
For simplicity I'll set $u = \begin{pmatrix}1 & -2 & -3 & -4\end{pmatrix}^T$ and $b = \begin{pmatrix}2&16&10&22\end{pmatrix}^T$.
First of all it should be clear that those are solutions, since $\lambda e_1 + \mu e_2\in K$.
Note that the solution could be written as $u + \kappa$, where $\kappa\in K$. Now we will see that all solutions are of this form.
Consider a vector not on that form $u+v$ where $v\notin K$. Now we have $M(u+v) = Mu + Mv = b + Mv$, but since $Mv\ne 0$ we have that $M(u+v)\ne b$ that is it's not a solution.
Now assume that there is a solution that is not on this form, that is a solution $u+v$ where $v\notin K$. But then we have $M(u+v) = Mu+Mv = b+Mv$, but as $Mv\ne 0$ we have $M(u+v)\ne b$, that is $u+v$ is not a solution which concludes that all solutions are on the claimed form.