If the shortest distance between the lines 2x+6=3y−6=4z−6 and 3x−λ=4y−26=5z+26 is 6, then the square of the sum of all possible values of λ is:
Answer: 384
Solutions · 1
Official solution
\[(2,3,4)\times(3,4,5)=(-1,2,-1),\qquad |(-1,2,-1)|=\sqrt{6}.\]
Vector between points: (6+λ,6−26,6−(−26))=(6+λ,−6,36); dot with (−1,2,−1) divided by 6 equals ±6:
\[\frac{1}{\sqrt{6}}\left|-(\sqrt{6}+\lambda)-2\sqrt{6}-3\sqrt{6}\right|=6 \implies \lambda=-2\sqrt{6}\ \text{or}\ 10\sqrt{6}.\]
Sum =86; square of sum =384.