Given the sphere

$\displaystyle (x-3)^2 + (y-4)^2 + z^2 = 36$

And the plane

$\displaystyle 3x-2y+4z=8$

Find the shorterst distance from the center of the sphere to the plane.

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This is what I've tried:

Center of the sphere $\displaystyle C(3, 4, 0)$

The normal vector of the plane $\displaystyle \vec n =[3, -2, 4]$

I chose an arbitrary point on the plane $\displaystyle P(x, y, z)$

Which gives us;

$\displaystyle [x, y, z]=[3, 4, 0]+[3, -2, 4]t$

So the distance should be

$\displaystyle \rm{D} = \sqrt{(3+3t)^2 + (4-2t)^2 + (4t)^2}$

Where;

$\displaystyle f(t) = 29t^2 + 2t + 25$

Which gives;

$\displaystyle f\prime(t) = 58t+2$

And

$\displaystyle f\prime(t) = 0$

$\displaystyle t = -\frac{2}{58}=\underline{-\frac{1}{29}}$

So

$\displaystyle f(-\frac{1}{29}) = 29(-\frac{1}{29})^2 + 2(-\frac{1}{29}) + 25 = \underline{\frac{724}{29}}$

$\displaystyle \rm{D} = \sqrt{\frac{724}{29}}$

But this is not correct according to my book. What am I doing wrong?