Answer
$(x - 1)^2+(y -2)^2+(z - 3)^2=14$
Work Step by Step
We know that the standard equation of a sphere is written as:
$(x-x_0)^2+(y-y_0)^2+(z-z_0)^2=r^2$ ...(1)
Here, $(x_0,y_0,z_0)$ represents the center and $r$ is the radius of the sphere.
Since, we have $x_0=1,y_0=2,z_0=3, r=\sqrt{14}$
Plug all the above values in equation (1), then we get
Thus,, $(x - 1)^2+(y -2)^2+(z - 3)^2=(\sqrt{14})^2$
Hence, $(x - 1)^2+(y -2)^2+(z - 3)^2=14$
