Use an intersection to determine an unknown coordinate

$AB=(4,6,\alpha-26)$, so line AB is $\mathbf r=(9,3,26)+s(4,6,\alpha-26)$, $s\in\mathbb R$.

$1+2t=9+4s$, $3-t=3+6s$ and $7+5t=26+s(\alpha-26)$. The first two give $-4s+2t=8$, $6s+t=0$, hence s=−1/2 and t=3. Substituting into the third yields $22=26-\frac12(\alpha-26)$, so α=34. The intersection is P=(7,0,22).

The directions are $(2,-1,5)$ and $(4,6,8)$, with squared magnitudes 30 and 116. Their dot product is 42. Thus

$\begin{aligned}&\theta\\ & =\cos^{-1}\left(\frac{|42|}{\sqrt{30}\sqrt{116}}\right)\\ & =44.6^\circ.\end{aligned}$

The absolute value explicitly selects the acute line angle; the printed positive numerator already gives this angle.