Answer
$(x+3)^2+(y-2)^2+(z-5)^2=4^2$
The sphere intersects the yz-plane at $(y-2)^2+(z-5)^2=4^2-3^2=7$
Work Step by Step
The equation can be found using the formula for a sphere:
$(x-h)^2+(y-k)^2+(z-l)^2=r^2$
in which the point $(h,k,l)$ is the center of the sphere, and $r$ is the radius of the sphere.
The intersection with the yz-plane can be determined by setting $x=0$
$(0+3)^2+(y-2)^2+(z-5)^2=4^2$
$(3)^2+(y-2)^2+(z-5)^2=4^2$
$(y-2)^2+(z-5)^2=4^2-3^2=7$
