Linear Diophantine equation with three variables and a condition

Let me start by saying that I'm new to Diophantine equations and my method certainly will not be the best possible one. I'm aware that a faster method of solving this particular case exists, but I want to know if what I did was correct and how I can finish the exercise.

$39x+55y+70z=3274$

$x+y+z=69$

$x\ge 0,y\ge 0,z\ge 0$

What I tried:

$39x+55y=3274-70z$

The GCD for 39 and 55 is one, so we can set aside $z=t$ to be a parameter, where t is an integer. We proceed to solve this as a Diophantine equation with two variables.

I now need to use the extended Euclidean algorithm to express 1 as a linear combination of 39 and 55.

I got that $1=24\cdot 39-17\cdot 55$

Now, the solution would be

$x=78576-1680t+55s$

$y=-55658+1190t-39s$

$z=t$

where t and s are integers. I got this from the formula that $x={\lambda}_{1}\cdot b+s\cdot {a}_{2}$ and $y={\lambda}_{2}\cdot b-s\cdot {a}_{1}$.

Where ${\lambda}_{1}$ and ${\lambda}_{2}$ are the coefficients in the Euclidean algorithm, b is $3274-70t$ and ${a}_{1}$ and ${a}_{2}$ are 39 and 55, respectively.

It's obvious that $t\ge 0$, and if I put that x and y are greater than zero, I get that

$s\ge \frac{1680t-78576}{55}$ and $s\le \frac{-55658+1190t}{39}$ which makes $\frac{1680t-78576}{55}\le \frac{-55658+1190t}{39}$ which is $t\le 46.77$.

I now have the whole range of integers [0,46] which of course isn't feasible to do (as I'd have to plug them into the inequalities for s and get even more cases.

How do I proceed from here? What do I do? I still have the condition $x+y+z=69$ but I feel like I'm missing something. Can it be really done this way, except that it's an unimaginably hefty job of testing each case?