Find parametric equations for the line through the point (0, 3, 2) that is parallel to the plane x + y + z = 5 and perpendicular to the line x = 1 + t, y = 3 − t, z = 2t. (use the parameter t.)

Hello : 
let A(0,3,2) and (Δ) this line , v vector   parallel to (Δ).
M∈ (Δ) : vector (AM) = t v t ∈ R

1 )    (Δ)  parallel to the plane x + y + z = 5 : let  : n an vector perpendicular 
to the plane : n ⊥ v      n(1,1,1) so : n.v =0  means : n.vector (AM) = 0
(1)(x)+(1)(y-3)+(1)(z -2) =0      ( vector (AM) = ( x, y -3 , z-2 )
x+y+z - 5=0 ...(1)

2)  (Δ) perpendicular to the line (Δ') : x = 1+t  , y = 3 - t , z = 2t :
vector (u) ⊥ v      vector(u) parallel to (Δ')  and vector(u) = (1 , -1 ,1)
vector (u) ⊥ vector (AM) means : 
(1)(x)+(-1)(y-3)+(2)(z -2) =0
x - y+2z - 1 = 0 ...(2)
so the system : 
x+y+z - 5=0 ...(1)
x - y+2z - 1 = 0 ...(2)
 (1)+(2) :   2x+3z - 6 =0
x = 3 - (3/2)z
subsct in (1) :    3 - (3/2)z  +y +z - 5 =0
y = 1/2z +2
let : z=t     
an parametric equations for the line (Δ) is :  x = 3 - (3/2)t
                                                                      y = (1/2)t +2
                                                                      z=t

verifiy : 
1) (Δ)  parallel to the plane x + y + z = 5 : 
(-3/2 , 1/2 ,1) perpendicular to (1,1,1)
because : (1)(-3/2)+(1)(1/2)+(1)(1) = -1 +1 = 0
2) (Δ) perpendicular to the line (Δ') :
 (-3/2 , 1/2 ,1) perpendicular to (1,-1,2)
because : (1)(-3/2)+(-1)(1/2)+(1)(2) = -2 +2 = 0
A(0, 3, 2)∈(Δ) : 
0 = 3-(3/2)t
3 = (1/2)t+2
2 =t
same :  t = 2

Use a calculator basically