Find the volume of the tetrahedron with vertices at $(0,0,0),(a,0,0),(0,b,0),(0,0,c)$

The most straight forward approach would be evaluating the following integral in Cartesian coordinates system.

$$V=int_0^aint^{b(1-x/a)}_{0}int^{c(1-x/a-y/b)}_0dzdydx=frac{abc}{6}$$

(which I can make sense of.)

But I tried two alternative methods each of which involves parametric equations:

Describing the surface that joins the three vertices, with a vector equation, I got

$$vec{r}=(a,0,0)+lambda(-a,b,0)+mu(-a,0,c)=(a(1-lambda-mu),lambda b,mu c)=(x,y,z) tag{1}$$

(here I used $vec{AB}=(-a,b,0)$ and $vec{AC}=(-a,0,c)$ as the two directional vectors.)

and

$$x=a(1-lambda-mu),y=lambda b, z=mu c tag{2}$$

as the three parametric equations that describe the surface, where $0leq yleq b$ and $0leq zleq c$ give the limits $0leq lambda leq 1$ and $0leq mu leq 1$.

Taking $dV=xdydz$:

$$dV=x(lambda,mu)frac{dy}{dlambda}frac{dz}{dmu}dlambda dmu=abc(1-lambda-mu)dlambda dmu tag{3}$$

Integration yields $$V=abcint^{mu=1}_{mu=0}int^{lambda=1}_{lambda=0}(1-lambda-mu)dlambda dmu=0 tag{4}$$

which is clearly wrong.

Alternatively, by taking $dV=dxdydz$, I got some rather dodgy differential terms:

$$dV=left(frac{partial x}{partial lambda}dlambda+frac{partial x}{partial mu}dmuright)frac{dy}{dlambda}frac{dz}{dmu}dlambda dmu=-abc(dlambda+dmu)dlambda dmu tag{5}$$

which has terms like $dlambda^2$ (the square of a differential is not something I’ve seen before, except for $ds^2=g_{munu}dx^{mu}dx^{nu}$ but that’s a different story…).

Can someone please explain where my conceptual errors lie?

Or is there a correct approach involving parametric equations?