Find the equation of the plane such that image of point `(1,2,3)` in it is`(-1,0,1)dot`

Explanation

Since the image of `A(1, 2, 3)` in the plane is `B(-1, 0, 1)`, the passes through the midpoint (0, 1, 2) of `AB` and is normal to the vector `vec(AB)-=-2hati-2hatj-2hatk`.
Hence, the equation of the plane is `-2(x-0)-2(y-0)-3(z-2)=0 or x+y=z=3.`