<s>[INST] Distance: sphere related question Three mutually tangent spheres of radius 1 rest on a horizontal plane. A sphere of radius $2$ rests on them. What is the distance from the plane to the top of the larger sphere ? 
How do we do here as I have four options as answer : 
(a) $3+\dfrac{\sqrt{30}}{2}$ 
(b) $3+\dfrac{\sqrt{69}}{3}$  
(c) $3+\dfrac{\sqrt{123}}{4}$ 
(d) $\dfrac{52}{9}$
 [/INST] Let the centers of your 3 spheres of radius 1 be $C_1, C_2, C_3$. Let the center of the sphere of radius 2 be $C_4$. These 4 points forms a tetrahedron with equilateral triangle base of length 2 ( $C_1C_2 = C_2C_3=C_3C_1=2$) , and a side length of 3 ($C_4C_1 = C_4C_2 = C_4C_3 = 3$). Let $O$ be the center of the equilateral base, then $OC_1 = \frac {2}{3} \sqrt{3} $.
Consider right triangle $OC_1 C_4$, we get that $OC_4 = \sqrt{ C_1C_4 ^2 - O C_1 ^2 } = \sqrt{ 9 - \frac {4}{3} } = \sqrt{ \frac {23}{3} } = \frac {\sqrt{69} }{3}$.
Hence, the distance from the plane to the top is $ 1 + \frac { \sqrt{69} }{3} + 2$.
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