Find the point on the graph of y= e^x at which the curvature is the greatest

Please see the step-by-step solution in the attached file.

3. Let P(x, y, z) be a generic point on the surface of revolution. Fix a point Q(0, y1, z) on the parabola with the same z-coordinate as P. Then we have
z = y1^2 (1)
The square of the distance from Q to the z-axis is y1^2, and the square of the distance from P to the z-axis is x^2 + y^2. As the two distances should be the same, we have y1^2 = x^2 + y^2, So substitute the above equation into (1)
z^2 = x^2 + y^2 (2)
(2) in the cylindrical coordinate is
z = r^2

4. The plane passes through point P (1, 3, 5), so the equation of the plane can be written as
a(x -1) + b(y - 3) + c(z - 5) = 0
The direction vector for the line is u =(5, 5, -2).
Now we pick another point on the line, let's say when t=0, the point Q is Q(0, 6, 3).
Then the vector PQ = (1-0, 3-6, 5-3) is perependicular to the normal vector the plane.
Hence we can take
u x PQ = (4, 5, -2) x (1, -3, 2) = (4, -10, -17) as the nomral vector.
so the equation of the plane is
4(x -1) - 10(y-3) - 17(z - 5) = 0
4x - 10y - 17z = -111.