How do I show that a curve given by parametric equations are contained in a plane?

The general equation of a plane is
Ax+By+Cz+D=0
or x+by+cz+d=0 (after dividing by A).

If you can find values for b, c, d which allow you to substitute the parametric equations into this form, then the curve is contained in a plane. This usually involves solving 3 simultaneous linear equations. Conversely if you can show that this cannot be done (that the 3 equations have no solution), then the curve does not lie in a plane.

After substituting, the equation in powers of t has to hold for any value of t, or for all powers of t individually.

First Method: you can take t=+1 and t=-1 to eliminate the coefficient of z (c), leaving equations between b and d :
(1+3+2) + b(2-2+5) + c(1-1) +d = 0
or 6 +5b +d =0  (i).
(1-3+2) + b(2+2+5) + c(1-1) +d =0
or 9b +d =0  (ii).
Eqn (ii) shows that d=-9b. Substitution in (i) gives
6 + 5b - 9b =0
6 = 4b
b = 6/4 = 3/2.
Then d = -9b = -9*3/2 = -27/2.
Another equation is required to find a value for c. Choosing t=0 is easy:
1 + 2b + c +d =0   (iii)
Substitute for b and d :
1+3+c-27/2=0
c = 27/2 - 4 = 27/2 -8/2 = 19/2.
c = -2b -d -1 = -3 -27/2 -1 = -35/2.

Values for b, c, d can be found, so the curve lies in the plane x+by+cz+d=0.

Second Method: collect powers of t -
t^0:  1 + 2b + c +d =0  (iv)
t^1:  3  -2b  = 0    (v)
t^2:  2 + 5b - c = 0   (vi)
From (v)  b=3/2.
From (vi) c=2+5b=2+15/2=19/2.
From (iv) 1+3+19/2+d=0 so d=-27/2.

Again, (the same) values have been found for b, c, d so the curve lies in this plane.