I think its easiest to understand this if one has a minimal understanding of QFT. I'm not sure about your background knowledge but hopefully this isn't gibberish to you.
The QCD Lagrangian for massless quarks is given by,
\begin{equation}
{\cal L} = - g \sum_i \bar{\psi} _i A _\mu \gamma ^\mu \psi _i - \frac{1}{4} F _{ \mu \nu } F ^{ \mu \nu }
\end{equation}
where the fields are, $ A _\mu $ and $ \psi _i $. The only constant in the equation is the coupling constant, $g$. Therefore, we see that there is no single scale in the Lagrangian. Naively one would say that the theory is scale invariant.
However, there is a subtlety. We haven't fully specified the theory. We have yet to say what the value of the coupling constant is. The problem is that QFT causes the strength of an interaction to depend on the scale which its measured. Luckily, we know how to calculate how a coupling changes with scale (this is done in every full year QFT course),
\begin{equation}
\frac{ d \alpha }{ d \log \mu } = - \frac{ b }{ 2\pi } \alpha ^2
\end{equation}
where, $ \alpha \equiv g ^2/4 \pi $ and $ b $ are calculable numbers. For QCD with the SM fermions we have,
\begin{equation}
b = 7
\end{equation}
From here its easy to solve the differential equation above and get the coupling as a function of the scale, $ \mu $,
\begin{align}
\frac{1}{ \alpha ( \mu ) } &= \frac{1}{ \alpha ( \mu _0 ) } - \frac{ b }{ 2\pi } \log \frac{ \mu }{ \mu _0 } \\
\alpha_s (\mu) &= \frac{ \alpha _s ( \mu _0 ) }{ 1 + \alpha _s ( \mu _0 ) \frac{ b }{ 2\pi } \log \frac{ \mu }{ \mu _0 } }
\end{align}
Therefore, we can measure the coupling at some scale and then know what it is at every scale. As pointed out by the OP, we can already see breaking of scale invariance since the couplings depend on scale.
Now we move on to the relation to $ \Lambda_{QCD} $. This is conventionally defined as the scale where the coupling becomes infinite. From the running above we see this occurs when,
\begin{equation}
\mu \equiv \Lambda_{QCD} = \mu _0\exp \left[ - \frac{ 2\pi }{ b \alpha _s ( \mu _0 ) } \right]
\end{equation}
Here we see that the scale only depends the field content (through $b$) and Natures choice for the coupling.
