where the function we want to minimize is actually [tex]\sqrt{x^2+(y-2)^2+(z-5)^2}[/tex], but it's easy to see that [tex]\sqrt{f(\mathbf x)}[/tex] and [tex]f(\mathbf x)[/tex] have critical points at the same vector [tex]\mathbf x[/tex].
Solving for [tex]x,y,z[/tex], we get a single critical point at [tex]\left(-\dfrac5{14},\dfrac{19}7,\dfrac{55}{14}\right)[/tex], which in turn gives the least distance between the plane and (0, 2, 5) of [tex]\dfrac5{\sqrt{14}}[/tex].