20. I suppose you meant to write [tex]3x-3y+2z=5[/tex]? This plane has normal vector (3, -3, 2), so the equation of the plane parallel to this one that goes through (4, 3, 5) is
[tex](3,-3,2)\cdot(x-4,y-3,z-5)=0\implies3(x-4)-3(y-3)+2(z-5)=0[/tex]
[tex]\implies 3x-3y+2z=13[/tex]
21. The tangent to [tex]x(t)[/tex] is parallel to the normal vector to the plane, so we can simply use it as the normal vector. This tangent vector is
[tex]\dfrac{\mathrm dx(t)}{\mathrm dt}=(1,-2,-1)[/tex]
Then the plane has equation
[tex](1,-2,-1)\cdot(x-1,y,z+3)=0\implies(x-1)-2y-(z+3)=0[/tex]
[tex]\implies x-2y-z=4[/tex]
