You need to apply the general method for solving a linear modular equation. Namely,
Existence $\,\ ax\equiv b\pmod{m} \iff ax\!-\!cm = b\iff (a,m)\mid b.\,$ If so, $\ b = (a,m)e\,$
hence Bezout $\,\Rightarrow\, \exists\, j,k\!:\ aj\!-\!km = (a,m)\,\stackrel{\times\, e}\Rightarrow\,a(je) -(ke)m = (a,m)e = b.$
Uniqueness $\,\ x\,$ is unique mod $\,m/d,\, \, d = (m,a),\,$ by $\,\color{#c00}{(m/d,a/d)=1}\,$ so by Euclid's Lemma
mod $\,m\!:\ ax\equiv b\equiv ax'\Rightarrow\, m\mid a(x\!-\!a')\,\Rightarrow\,m/d\mid a/d(x\!-\!x')\,\smash[t]{\stackrel{\rm\color{#c00}{Euclid}}\Rightarrow}\, m/d\mid x\!-\!x'$
Remark $\ $ You can locate lterature on thus and related elimination techniques by searching on Hermite normal form (or Smith normal form).
In part (a), we find that every solution to $A\vec{x}=\vec{O}$ is of the form
\begin{align*}
\vec{x}
&= \left[\begin{array}{r}
x_{1} \\
x_{2} \\
x_{3} \\
x_{4} \\
x_{5} \\
x_{6}
\end{array}\right]
= \left[\begin{array}{r}
-2 \, x_{2} - 3 \, x_{4} - 5 \, x_{6} \\
x_{2} \\
-4 \, x_{4} - 2 \, x_{6} \\
x_{4} \\
3 \, x_{6} \\
x_{6}
\end{array}\right] \\
&= x_2\left[\begin{array}{r}
-2 \\
1 \\
0 \\
0 \\
0 \\
0
\end{array}\right]+x_4\left[\begin{array}{r}
-3 \\
0 \\
-4 \\
1 \\
0 \\
0
\end{array}\right]+x_6\left[\begin{array}{r}
-5 \\
0 \\
-2 \\
0 \\
3 \\
1
\end{array}\right]
\end{align*}
This means that the three vectors
\begin{align*}
\vec{v}_1 &= \left\langle-2,\,1,\,0,\,0,\,0,\,0\right\rangle &
\vec{v}_2 &= \left\langle-3,\,0,\,-4,\,1,\,0,\,0\right\rangle &
\vec{v}_3 &= \left\langle-3,\,0,\,-4,\,1,\,0,\,0\right\rangle
\end{align*}
form a basis of $\operatorname{Null}(A)$.
In particular, the vector $\vec{v}_1=\left\langle-2,\,1,\,0,\,0,\,0,\,0\right\rangle$ satisfies $A\vec{v}_1=\vec{O}$. This means that
$$
(-2)\cdot\vec{a}_1+1\cdot\vec{a}_2+0\cdot\vec{a}_3+0\cdot\vec{a}_4+0\cdot\vec{a}_5+0\cdot\vec{a}_6=\vec{O}
$$
where $\{\vec{a}_1, \vec{a}_2, \vec{a}_3, \vec{a}_4, \vec{a}_5, \vec{a}_6 \}$ are the columns of $A$. This gives $\vec{a}_2=2\cdot\vec{a}_1$.
So, if the first column of $A$ is $\vec{a}_1=\left\langle2,\,3,\,5\right\rangle$, then the second column of $A$ is $\vec{a}_2=2\cdot\vec{a}_1=\left\langle4,\,6,\,10\right\rangle$.
Also note that the vector $\vec{v}_2=\left\langle-3,\,0,\,-4,\,1,\,0,\,0\right\rangle$ satisfies $A\vec{v}_2=\vec{O}$. This means that
$$
3\cdot\vec{a}_1+0\cdot\vec{a}_2+(-4)\cdot\vec{a}_3+1\cdot\vec{a}_4+0\cdot\vec{a}_5+0\cdot\vec{a}_6=\vec{O}
$$
Solving for $\vec{a}_4$ gives
$$
\vec{a}_4 = -3\cdot\vec{a}_1+4\cdot\vec{a}_3
$$
So, if $\vec{a}_1=\left\langle2,\,3,\,5\right\rangle$ and $\vec{a}_3=\left\langle3,\,-1,\,4\right\rangle$, then
$$
\vec{a}_4
= -3\cdot\vec{a}_1+4\cdot\vec{a}_3
= -3\cdot\left\langle2,\,3,\,5\right\rangle+4\cdot\left\langle3,\,-1,\,4\right\rangle
= \left\langle6,\,-13,\,1\right\rangle
$$
Best Answer
Remember that augmented matrices correspond to systems of linear equations. Once you've finished row-reducing, turn the row-reduced matrix back into a system of equations and solve for the variables in the pivot columns:
$$\begin{pmatrix}1 & 6 & 0 & 11 & | & 0 \\ 0 & 0 & 1 & -8 & | & 0 \\ 0 & 0 & 0 & 0 & | & 0\end{pmatrix} \longrightarrow \begin{cases}x_1 + 6x_2 + 11x_4 = 0 \\ x_3 -8x_4 = 0\end{cases}\longrightarrow \begin{cases}x_1 = -6x_2 - 11x_4 \\ x_3 = 8x_4.\end{cases}$$ The free variables $x_2,x_4$ are now parameters. Once you specify them, you specify a single solution to the equation. So subsitute $x_2 = s,x_4 = t$ and arrive at the parametrized form: $$\begin{cases} x_1 = -6s - 11t\\ x_2 = s\\ x_3 = 8t\\ x_4 = t \end{cases}$$