Find the volume of the solid bounded in the first octant of the coordinate planes and by the plane 3x+4y+2z=12

The volume of the solid bounded in the first octant of the coordinate planes and by the plane 3x+4y+2z=12 can be found by using triple integrals. The limits of integration for x, y, and z can be found by setting two of the variables equal to zero and solving for the third variable. For example, when x=0 and y=0, we have 2z=12, so z=6. Similarly, when x=0 and z=0, we have 4y=12, so y=3. And when y=0 and z=0, we have 3x=12, so x=4. Therefore, the limits of integration are 0≤x≤4, 0≤y≤3-3x/4, and 0≤z≤6-3x/2-2y. The triple integral that gives the volume of the solid is: $$ \int_0^4\int_0^{\left(3-\frac{3}{4}x\right)}\int_0^{\left(6-\frac{3}{2}x-2y\right)} $$dzdydx Evaluating this integral gives us a volume of 8 cubic units. So, the volume of the solid bounded in the first octant of the coordinate planes and by the plane 3x+4y+2z=12 is 8 cubic units.