Triple integral tetrahedron

Hi everyone I am having trouble what this question is asking me to do. Is it just asking me to write the initial equation? If so I have a rough idea how to do this for double integrals but not so for triple. Could you please solve the question and show the steps so I get where I'm going wrong. Thanks for the help

Formulate a triple integral to compute the mass of an object
which is in the shape of the tetrahedron bounded by the planes x = 0,
y = 0, z = 0 and z = 6 - 3x -2y and whose density is given by the
function f(x; y; z). You do not have to actually evaluate the integral.

Finding the domain.

x = 0 , y = 0 , z = 0 mean the first octant ;

z varies :
0 ≤ z ≤ 6 - 3x - 2y

For z = 0 ( in the plane xOy ) get y = - 3x/2 + 3 , so
0 ≤ y ≤ - 3x/2 + 3

And x ( (y=0 , z=0) = axis x ) varies :
0 ≤ x ≤ 2

the domain
D = {(x,y,z) | 0 ≤ x ≤ 2 ; 0 ≤ y ≤ - 3x/2 + 3 ; 0 ≤ z ≤ 6 - 3x - 2y }

The mass :
M = ∫(x = 0 to 2) ∫(y = 0 to 3 - 3x/2) ∫(z = 0 to 6 - 3x - 2y) f(x,y,z) dz dy dx

Of course x ≥ 0 , y ≥ 0 , z ≥ 0 mean the first octant !
My hand is slapped.

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