Ka-Boom!

The simplified form of the question is

$3x+4y+5z=20$ . Find least value of $x^{2}+y^{2}+z^{2}$

The first equation represents the equation of a plane. The second expression is just the square of distance from the origin to any point on the plane.

Therefore the question is simply to find the minimum distance of the plane of the origin and then square it.

By using basic 3D geometry we can easily find the minimum distance between plane and the origin.

For a genereal equation $ax+by+cz=d$ the minimum distance is $\frac{d}{\sqrt{a^{2}+b^{2}+c^{2}}}$ . Therefore the distance squared will be $[\frac{d}{\sqrt{a^{2}+b^{2}+c^{2}}}]^{2}$ .

Putting $a=3,b=4,c=5,d=20$ in the above expression we get the value $\boxed{8}$

By Cauchy-Schwarz we have $20^2\leq (3^2+4^2+5^2)(\tan^2x+\tan^2y+\tan^2z)$ so $\tan^2x+\tan^2y+\tan^2z\geq \boxed{8}$ . Equality is attained when $\tan{x}=1.2, \tan{y}=1.6, \tan{z}=2$ .