The parametric equation ofa line typically takes the form $\underline r=\underline pt+ \underline q$,
where $\underline p$ is the "direction vector" of the line and $\underline q$ is the position vector of a known point on the line.
In the situation you describe, the line $z=z_0+a(x-x_0)$ has gradient $a$ and passes through the point $(x_0, y_0, z_)$.
Its parametric or vector equation is $\underline r=\left(\begin{array}{r}
1 \\
0 \\
a
\end{array}\right)t + \left(\begin{array}{r}
x_0 \\
y_0 \\
z_0
\end{array}\right)
$
Similarly, the line $z=z_0+b(y-y_0)$ has gradient $b$ and passes through the point $(x_0, y_0, z_)$.
Its parametric or vector equation is $\underline r=\left(\begin{array}{r}
0 \\
1 \\
b
\end{array}\right)t + \left(\begin{array}{r}
x_0 \\
y_0 \\
z_0
\end{array}\right)
$
The vector product of the two direction vectors is $\underline n=\left(\begin{array}{r}
1 \\
0 \\
a
\end{array}\right) \times
\left(\begin{array}{r}
0 \\
1 \\
b
\end{array}\right)=\left(\begin{array}{r}
-a \\
-b \\
1
\end{array}\right)
$
The equation of the plane is given by $\underline r .\underline n = c$
Substitute $\underline n=\left(\begin{array}{r}
-a \\
-b \\
1
\end{array}\right)
$ and the known point $\underline r=\left(\begin{array}{r}
x_0 \\
y_0 \\
z_0
\end{array}\right)
$ to get $c=z_0-ax_0-by_0$
