We have $$\iiint_{H}(x+y+z)^{2} d V$$
where $H$ is the region:
$0 \leq x+y+z \leq 1, \quad 0 \leq x+2 z \leq 3, \quad 0 \leq x+2 y+3 z \leq 2$
By use of the change of variables, $u=x+y+z, \quad v=x+2 z, \quad w=x+2 y+3 z$.
Apparently, the answer is $$\frac{\partial(u, v, w)}{\partial(x, y, z)}=\left|\begin{array}{ccc}{1} & {1} & {1} \\ {1} & {0} & {2} \\ {1} & {2} & {3}\end{array}\right|=-3$$
$$\begin{array}{c}{\iiint_{H}=\int_{0}^{2} \int_{0}^{3} \int_{0}^{1} u^{2}\left|\frac{\partial(x, y, z)}{\partial(u, v, w)}\right| d u d v d w} \\ {=1 / 3 \int_{0}^{2} d w \int_{0}^{3} d v \int_{0}^{1} u^{2} d u} \\ {=1 / 3 \times 2 \times 3 \times 1 / 3=2 / 3}\end{array}$$
However, I don't understand how we can compute the determinant of the partials of u,v and w with respect to x,y and z instead of the other way round? When I tried to compute the partials the other way round and take the derivative i got $\frac{5}{12}$ which is not the reciprocal of 3.
This answer seems correct. You must have made a mistake when computing $\det \dfrac{\partial(x,y,z)}{\partial(u,v,w)}$ directly. The reason why we can compute $\det \dfrac{\partial(u,v,w)}{\partial(x,y,z)}$ instead is that these matrices are inverses of each other (if you're careful about where the derivatives are being evaluated); so knowing the determinant of one tells you that the determinant of the other one is the reciprocal.
You're right that what you actually need to compute is \begin{equation} \alpha := \det \dfrac{\partial(x,y,z)}{\partial(u,v,w)} \bigg|_{(u,v,w)} \end{equation} But, in this case, it is easier to compute \begin{equation} \det \dfrac{\partial(u,v,w)}{\partial(x,y,z)} \bigg|_{(x,y,z)} = -3 \end{equation} because we already have $u,v,w$ written in terms of $x,y,z$. From this, we can immediately deduce from properties of determinants and inverse matrices that $\alpha = -\dfrac{1}{3}$.