Determine the point of intersection of the plane with vector equation....

Determine the point of intersection of the plane with vector equation [5,3,-2] + s[1,3,2] + t[2,4,0] and the y-axis
I know I have to find the parametric equation and then find the value of s and t and solve.. but I don't know how to get it!
PLEASE help! and please show how to get the answer!

the y-axis coordinates is zero, only in the y-coordinate, then a point on the y axis is of the form (0, y 0). This point intersects the plane. Once this point satisfies the equation of the plane. Thus (0, y 0) = (5,3, -2) + s (1,3,2) + t (2,4,0)
so 1) 5 + s + 2t = 0
2) 3 + 3s + 4t = y
3) -2 + 2s +0 = 0 ,so s = 1
from 1) we have 5 +s + 2t = 5 +1 +2t = 0 , so t = -3
from 2) 3 + 3s + 4t = 3 +3*1 + 4*(-3) = -6 = y
(0,-6,0) Ans

x = 5 + s + 2 t ; y = 3 + 3s + 4t , z = - 2 + 2s.....y axis is ( 0 , y , 0 )

thus s = 1 , t = -3 , y = - 5