Solution
= Solution
Parametrize the first line by $(x,y,z)=(3+t,4-2t,1+3t)$ and the second by $(2+\alpha s,-1+s,5+2s)$. Equality of the $y,z$ components gives $s=5-2t$ and $3t-2s=4$, hence $t=2$, $s=1$. The $x$ components then require $5=2+\alpha$. Thus
$$
\boxed{\alpha=3,\qquad\text{intersection }(5,0,7).}
$$
The two independent component equations fixed the parameters before the remaining component determined $\alpha$, so no extra intersection case was discarded.