Find a common line and use normals for the acute plane angle

$x-y+3z=3$ and $2x-y+2z=5$ give $x=2+z$, $y=-1+4z$. Hence $\mathbf r=(2,-1,0)+\lambda(1,4,1)$, $\lambda\in\mathbb R$.

The normals have lengths $\sqrt{11}$ and $3$, and dot product $9$. The acute angle is $\cos^{-1}(9/(3\sqrt{11}))\approx0.441$ rad (3 s.f.), or $25.2^\circ$ (1 d.p.).

The normals are $(1,1,0)\times(1,0,1)=(1,-1,-1)$ and $(1,2,4)\times(0,1,1)=(-2,-1,1)$. Their cross product is $(-2,1,-3)$. Both planes contain $(1,2,7)$, so the intersection is $\mathbf r=(1,2,7)+\alpha(-2,1,-3)$.

The scalar equations are $x-y-z=-8$ and $-2x-y+z=3$. Solving for x,y in terms of z gives $x=-11/3+2z/3$, $y=13/3-z/3$. Thus an equivalent line is $\mathbf r=(-11/3,13/3,0)+\alpha(2/3,-1/3,1)$. At parameter z=7 it passes through $(1,2,7)$; its direction is $-1/3$ times the analytical direction.

The acute angle is $\cos^{-1}(|-2|/(\sqrt3\sqrt6))\approx1.08$ rad, or $61.9^\circ$. These displayed values are 3 s.f.; the printed source labels both as 5 s.f., which is a rounding-label error. The exact expression removes that ambiguity.