5. (15 pts) Find the volume of the tetrahedron enclosed by the coordinate planes and the plane $\quad x+y+z=3$.

Step 1 ：The volume of a tetrahedron is given by the formula \(\frac{1}{6} \times base \times height\). In this case, the base is a triangle formed by the intersection of the plane with the coordinate axes, and the height is the distance from the origin to the plane.

Step 2 ：The base triangle has vertices at (3,0,0), (0,3,0), and (0,0,3), so its area can be calculated using the formula for the area of a triangle, \(\frac{1}{2} \times base \times height\). The area of the base triangle is 4.5.

Step 3 ：The height of the tetrahedron is the distance from the origin to the plane, which can be calculated using the formula for the distance from a point to a plane, \(\frac{|Ax+By+Cz+D|}{\sqrt{A^2+B^2+C^2}}\), where (A,B,C) are the coefficients of the plane equation and (x,y,z) is the point. In this case, A=B=C=1, and D=-3, and the point is the origin, (0,0,0). The distance from the origin to the plane is approximately 1.7320508075688774.

Step 4 ：Substituting the area of the base triangle and the height of the tetrahedron into the formula for the volume of a tetrahedron, we get the volume of the tetrahedron is approximately 1.299038105676658.

Step 5 ：Final Answer: The volume of the tetrahedron enclosed by the coordinate planes and the plane \(x+y+z=3\) is \(\boxed{1.299}\).