[tex]-3x-5y+3z=3[/tex] has normal vector (-3, -5, 3)
[tex]-5x+3y-5z=-4[/tex] has normal vector (-5, 3, -5)
Compute the angle between these normal vectors:
[tex](-3,-5,3)\cdot(-5,3,-5)=\|(-3,-5,3)\|\|(-5,3,-5)\|\cos\theta\implies\theta\approx1.873\,\mathrm{rad}\approx107.326^\circ[/tex]
