Look down onto the $y,z$ plane. In the plot below, the front wall of the bounding frame corresponds to $x=0$ and the back wall to $x=12$.
You'll notice that the two surfaces $x=y$ and $x=3z^2$ (resp. blue and orange) compete for "bottom surface" with respect to the $x$-axis. They intersect in a parabolic curve (red) along the wall of the parabolic cylinder. To one side of this parabola (below it, from this angle) the plane $x=y$ lies beneath the cylinder $x=3z^2$, i.e. $x=y$ is closer to $x=0$ and $x=3z^2$ is closer to $x=12$; in this case, the cylinder acts as the floor for the given region. Otherwise, to the other side of the parabola ("above" it) $x=3z^2$ lies below $x=y$, so the latter is closer to $x=12$ and acts as lower bound.
Over the whole rectangle $(y,z)\in[0,12]\times[-2,2]$, we can then say $x$ must be between $12$ and the larger of the two functions $x=y$ and $x=3z^2$, denoted $\max\left\{y,3z^2\right\}$.
So we can capture the whole region with the condensed triple integral,
$$\int_{z=-2}^2 \int_{y=0}^{12} \int_{x=\max\left\{y,3z^2\right\}}^{12} f(x,y,z) \, dx \, dy \, dz$$
The OP only mentions a generic integral, not a volume, so I assume $f\neq1$. We need to carefully split up the domain if we actually want to compute this.
For simplicity, let's suppose we just want the volume and that $f=1$. For comparison, using the much simpler order $dy\,dx\,dz$, we have a volume of
$$\begin{align*}
I_{yxz} &= \int_{z=-2}^2 \int_{x=3z^2}^{12} \int_{y=0}^x dy \, dx \, dz \\
&= 2 \int_0^2 \int_{3z^2}^{12} \int_0^x dy \, dx \, dz \\
&= \frac{1152}5
\end{align*}$$
Now for our target order $dz\,dy\,dx$, we must split up the $y$ integral depending on which subset of the region we find ourselves in. For reference, consult the plot below showing a projection of the region's boundaries onto the $y,z$ plane.
$$\begin{align*}
I_{xyz} &= \int_{z=-2}^2 \int_{y=0}^{12} \int_{x=\max\left\{y,3z^2\right\}}^{12} dx \, dy \, dz \\
&= 2 \int_0^2 \int_0^{12} \int_{\max\left\{y,3z^2\right\}}^{12} dx \, dy \, dz \\
&= 2 \left(\int_0^2 \int_0^{3z^2} \int_{3z^2}^{12} dx \, dy \, dz + \int_0^2 \int_{3z^2}^{12} \int_y^{12} dx \, dy \, dz\right) \\
&\stackrel{\checkmark}= \frac{1152}5
\end{align*}$$
