Find a line intersection and its acute angle

The printed answer is (3,6,13). To obtain it, equate $1+\lambda=2+\mu$, $2+2\lambda=1+5\mu$, $3+5\lambda=7+6\mu$. The first gives λ=1+μ; substituting into the second gives $4+2\mu=1+5\mu$, so μ=1 and λ=2. Check the third: $3+5(2)=7+6(1)=13$. The common point is $(1,2,3)+2(1,2,5)=(3,6,13)$.

The dot product is $(1,2,5)\cdot(1,5,6)=1+10+30=41$, with magnitudes $\sqrt{30}$ and $\sqrt{62}$. Thus $\alpha=\cos^{-1}(41/\sqrt{1860})=18.1^\circ$ to 1 decimal place, or 0.315 rad to 3 significant figures, matching both printed answers. These derivations supplement the source answer list.