Please find the solution to the following systems:

c2 + 6c3 = 8
3c2 + c3 = 3
4c2 + (-2)c3 = 8

I couldn't find any solution that satisfies all three, so you can't
write u = {8,3,8} as a linear combo of vectors, right?

Sorry for my broken English! I shall try to help you.

Your system is equivalent to the system (*):
c2 + 6c3 = 8
3c2 + c3 = 3
(4c2 + (-2)c3 = c2 + 6c3) <=> (3c2 = 8c3) <=> (c2 = 8c3/3) .

Hence the last system is equivalent to the system (**):
8c3/3 + 6c3 = 8
3(8c3/3) + c3 = 3
c2 = 8c3/3 .

But the system (***):
(8c3/3 + 6c3 = 8)   <=> (c3 = 12/13)
(3(8c3/3) + c3 = 3) <=> (c3 = 1/3)
has not a solution, because 12/13 =/= 1/3 .

Consequently the system (**) also has not a solution, because the
system (***) is the subsystem of the system (**). And consequently the
system (*) also has not a solution since the system (*) is equivalent
to the sytem (**). Similary the original system ( It is equivalent to
sytem (*) ) has not a solution.

Answer.
There are no possible solutions to your system!
You can't write 
u = {8,3,8} 
as a linear combo of vectors:
u = {c2,3c2,4c2} + {6c3,c3,(-2)c3}.
You are right!

Are you sure there is no "c1" in any of the equations, so that this is
a system of 3 equations in 2 variables, c2 and c3?

dikun-ga's conclusion is more or less correct, though the final
paragraph should say that you can't write u = (8,3,8) as a linear
combo of vectors (1,3,4) and (6,1,-2).

An easier way of doing it without matrix arithmetic is as follows:
Take three times the first equation:  3c2 + 18c3 = 24
and subtract the second equation:     3c2 +   c3 =  3
This gives 17c3 = 21, so c3 = 21/17. The first equation then gives c2
=  8 - 6*21/17 = 10/17. You can check that these values satisfy the
first two equations. Then if we put them into the third equation we
get 40/17 - 42/17 = 8, which is clearly false. (You can do the same
thing using a different pair of equations to start with; you'll get a
different set of values for c2 and c3, which won't work when applied
to the remaining equation.)

Given the nature of the question, though, I'd say you're probably
expected to solve it by setting it up as a matrix and using Gaussian
elimination to put it in row-echelon form. This goes as follows (I've
only done one elementary row operation per step; at a more advanced
level you'd normally combine some of these):

[ 1  6 | 8 ]   [ 1    6  |   8 ]
[ 3  1 | 3 ] ~ [ 0  -17  | -21 ] (R2 <- R2 - 3*R1)
[ 4 -2 | 8 ]   [ 4   -2  |   8 ]

    [ 1    6 |   8 ]
 ~  [ 0  -17 | -21 ] (R3 <- R3 - 4*R1)
    [ 0  -26 | -24 ]

    [ 1    6 |   8   ]
 ~  [ 0    1 | 21/17 ] (R2 <- -R2/17)
    [ 0  -26 |  -24  ]

    [ 1    6 |    8   ]
 ~  [ 0    1 |  21/17 ] (R3 <- R3 + 26*R2)
    [ 0    0 | 138/17 ] 

The last line then says 0c1 + 0c2 = 138/17, i.e. 0 = 138/17, so there
is no solution.