Extract the normal vectors from the given planes:
[tex]-2x+y+z+2=0\implies\vec n_1=(-2,1,1)[/tex]
[tex]x+y-3z+1=0\implies\vec n_2=(1,1,-3)[/tex]
(which are unique up to their signs, meaning either [tex]\vec n_1[/tex] or [tex]-\vec n_1[/tex] are valid choices for the normal vector)
The third plane must be perpendicular to both these given planes, which means it would be parallel to both [tex]\vec n_1[/tex] and [tex]\vec n_2[/tex], which in turn means its own normal vector [tex]\vec n_3[/tex] should be perpendicular to both [tex]\vec n_1[/tex] and [tex]\vec n_2[/tex].
Enter the cross product:
[tex]\vec n_3=\vec n_1\times\vec n_2=(-4,-5,-3)[/tex]
or (4, 5, 3), which also works.
The given plane passes through (-1, 1, 4), so its equation is
[tex](x+1,y-1,z-4)\cdot\vec n_3=0[/tex]
Simplify:
[tex](x+1,y-1,z-4)\cdot(4,5,3)=0[/tex]
[tex]4(x+1)+5(y-1)+3(z-4)=0[/tex]
[tex]\boxed{4x+5y+3z=13}[/tex]
