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**Hint:**We use the visual process to find the rhomboid pyramid which gives us the value of the intersecting points. We use the height to find the volume of the rhomboid pyramid. The tetrahedron will be $\dfrac{1}{4}$ volume of the pyramid. We find the volume of the tetrahedron.

**Complete step by step answer:**

We are going to apply the visualisation process to get the tetrahedron.

There are 4 planes involved. Three planes of the coordinates and the fourth one being $2x+y+z=4$.

The coordinate planes are $x-y,y-z,x-z$ planes. The plane $2x+y+z=4$ intersects these coordinate planes. It becomes a rhombic pyramid.

We first find the intersections.

We put the points on $2x+y+z=4$.

The intersection on the X-axis will be when $y=z=0$ and value of $x=2$.

The intersection on the Y-axis will be when $x=z=0$ and value of $y=2$.

The intersection on the Z-axis will be when $y=x=0$ and value of $z=2$.

The diagonal distance of those points will be equal to the value of the coordinates.

The points are $\left( 2,0,0 \right),\left( 0,4,0 \right),\left( 0,0,4 \right)$ and distances are $2,4,4$ units from $\left( 0,0,0 \right)$.

From the z intersection, we get the height of the hypothetical rhomboid pyramid.

From the x and y intersections, we get half of each diagonal distance across the hypothetical base.

Therefore, the height is 4 units. The area A of base will be $4\left( \dfrac{1}{2}xy \right)=2\times 2\times 4=16$.

Volume of the pyramid will be $\left( \dfrac{1}{3}Ah \right)=\dfrac{16\times 4}{3}$.

The tetrahedron will be $\dfrac{1}{4}$ volume of the pyramid. This gives $\dfrac{16\times 4}{3\times 4}=\dfrac{16}{3}$.

The volume of the tetrahedron is $\dfrac{16}{3}$.

**Note:**

We can also apply the calculus triple integral process to find the volume for the tetrahedron. We integrate along the three axes to find the volume.

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