Two paths represented by [x, z] = [-82.2, 43] +s [0.2, 1.7] and [x, y] =[-84.4, 45.5] +t [2.4 -0.9]. Determine the intersection of the two paths.

I assume both are in x, y and not three D or you would have shown 0 in y for the first and 0 in z for the second

first vector:

x = -82.2 + 0.2 s
y = +43.0 + 1.7 s

second vector:

x = -84.4 + 2.4 t
y = +45.5 - 0.9 t

do they hit?
-82.2 + .2 s = -84.4 + 2.4 t
or
2.4 t - .2 s = 2.2
and
43 + 1.7 s = 45.5 -.9 t
or
.9 t + 1.7 s = 2.5

multiply first equation by .9 and second by 2.4
2.16 t - .18 s = 1.98
2.16 t + 4.08 s = 6
------------------- subtract
- 4.26 s = - 4.02
s = .944
then find t
t = .995
then x = -82.2 + .1888 = -82.01
from the first
and
x = -82.01 from the second
and for y
y = 44.6 from the first
and
y = 44.6 from the second
so I think about at
(-82,+45 )