Determining time-invariance of a system
You have a system with the following input-output relation:

$$y(t)=\int_{-t}^{\infty}x(-3\tau)d\tau\tag{1}$$

In order to check whether the system is time-invariant or not, we need to compare the shifted output with the output resulting from a shifted input. The shifted output is

$$y(t-T)=\int_{-(t-T)}^{\infty}x(-3\tau)d\tau\tag{2}$$

Shifting the input means applying an input signal $x_T(t)=x(t-T)$. Note that $x_T(-3\tau)=x(-3\tau-T)$, so the response to the shifted input is

$$y_T(t)=\int_{-t}^{\infty}x(-3\tau-T)d\tau\tag{3}$$

The system is time-invariant if $(3)$ equals $(2)$. So we try to make the integrand in $(3)$ look like the one in $(2)$ by substituting $-3\tau-T$ by $-3\zeta$, which gives

$$y_T(t)=\int_{-(t-T/3)}^{\infty}x(-3\zeta)d\zeta\tag{4}$$

Now $(2)$ and $(4)$ have the same integrand, but we see that the lower integration limits are different. Consequently, $y_T(t)\neq y(t-T)$, hence the system is not time-invariant.