The intersection of the spheres $$x^2 + y^2 + z^2 + 7x - 2y - z = 13$$ and $$x^2 + y^2 + z^2 - 3x + 3y + 4z = 8$$ is the same as the intersection of one of the sphere and the plane

Let

$$S_1=x^2+y^2+z^2+7x-2y-z-13=0$$

and

$$S_2=x^2+y^2+z^2-3x+3y+4z-8=0$$

The common curve of intersection of two spheres lies in their radical plane.

Therefore, the required plane is obtained by subtracting the equations of the spheres.

$$S_1-S_2=0$$

$$7x-2y-z-13+3x-3y-4z+8=0$$

$$10x-5y-5z-5=0$$

Dividing by $$5$$,

$$2x-y-z-1=0$$

Hence, the common intersection of the two spheres is the same as the intersection of either sphere with the plane

$$\boxed{2x-y-z-1=0}$$