Changing limit of integral make time-variant to time-Invariant system
The system-1 whose input/output equation is
$$ y(t) = \mathcal{T}\{x(t)\} = \int_{-\infty}^{t} x(\tau) d\tau $$ is time-invariant.

Whereas the system-2
$$ y(t) = \mathcal{T}\{x(t)\} = \int_{-\infty}^{3t} x(\tau) d\tau $$ is time-varying.

On the other hand the system-3
$$ y(t) = \mathcal{T}\{x(t)\} = \int_{-\infty}^{3t-3t_0} x(\tau) d\tau $$ is still time-varying for any (fixed) value of $t_0$.

I presume that your teacher have tried to underline the fact that what makes the system-2 time-varying is the fact that $y(t-t_0) \neq \mathcal{T}\{x(t-t_0)\}$ which becomes as the equation shows:

$$y(t-t_0) = \int_{-\infty}^{3(t-t_0)} x(\tau) d\tau  = \int_{-\infty}^{3t-3t_0} x(\tau) d\tau  $$

and 
$$\mathcal{T}\{x(t-t_0)\} = \int_{-\infty}^{3t} x(\tau-t_0) d\tau = \int_{-\infty}^{3t-t_0} x(\tau) d\tau $$

These two would be equal if th upper limit of the second equation would be $3t-3 t_0$. That's what your teacher told. Since they are not equal therefore the system-2 is time-varying.

Note that artifically making the limit as $3t-3t_0$ is quite meaningless, as the fixed value of $t_0$ has nothing to do with the variable amount of shifts that occurs in $x(t-d)$ or $y(t-d)$.