Hi there having some trouble with this.
The plane through the three points $(5, 4, -8)$,$(1, 6,-3)$ and $(7,-2,5)$
so I then converted it to
$(5, 4, -8) + s(-4, 2, 5) + t(2, -6, 13)$
then converted got $x, y, z$
$x = 5 - 4s +2t$
$y = 4 + 2s - 6t$
$z = -8 + 5s + 13t$
Now I have trouble converting it I'm getting all kinds of fractions and stuff that can't be right can anyone help me out?
Hint: You are taking the long way around.
For a plane, you need only two pieces of information: a point on the plane (say its coordinate vector is $\vec{r}_0$) and a vector $\vec{n}$ which is normal to the plane. This being the case, the equation of the plane is that $\vec{r}$ is on the plane if and only if $$ \vec{n}\cdot(\vec{r}-\vec{r}_0)=0. $$ If we write $\vec{n}=\langle a,b,c\rangle$, $\vec{r}=\langle x,y,z\rangle$, and $\vec{r}_0=\langle x_0,y_0,z_0\rangle$, we get the familiar $$ a(x-x_0)+b(y-y_0)+c(z-z_0)=0. $$ In this case, there are plenty of points available; if you can find two vectors that must lie "along" the plane, then their cross product will be a good choice for a normal vector.
Alternatively, we can use the work you've already done. You already showed that the plane can be parameterized by $$ x=4-4s+2t\qquad y=4+2s-6t\qquad z=-8+5s+13t, $$ for all $s$ and $t$. We want to write $z=ax+by+c$; note that $$ -8+5s+13t=z=ax+by+c=a(4-4s+2t)+b(4+2s-6t)+c. $$ Rearranging this yields $$ -8+5s+13t=s(-4a+2b)+t(2a-6b)+(4a+4b+c). $$ So, we must have $-4a+2b=5$ and $2a-6b=13$, to get the coefficients of $s$ and $t$ to on the LHS and RHS to match. We must also have $4a+4b+c=-8$ to get the constants right. Adding twice the second equation to the first yields $-10b=31$, or $b=-\frac{31}{10}$. This then yields $a=-\frac{14}{5}$. Plugging both of these in to $4a+4b+c=-8$, we find $c=\frac{92}{5}$, so that an equation for the plane is $$ z=-\frac{14}{5}x-\frac{31}{10}y+\frac{92}{5}. $$ Multiplying both sides by 10 and rearranging yields $$ 28x+31y+10z=184, $$ which is a bit nicer to look at.
