Determine all three-digit positive integers *abc* Determine all three-digit positive integers abc (a is hundreds b tens and c ones) such that 8abc = 3cba.

Here is my how I have started:
8 (100a + 10b + c) = 3(100c + 10b + a)
800a + 80b + 8c = 300c + 30b + 2a
797a + 50b - 292c = 0
50b = 292c - 797a
b = (292c - 797a)/50
Since abc and cba both need to be even we know that a must be even but not c because factor 292 makes the product even.
Then I started to test different values for a and c, it didn't take so long because a can be larger than 2, so the only values I tested were 1 and 2 for a. For c I tested all values up to 7 (then I found the answer) and it was kind of easy because you don't need to do the whole calculation since if the unit digit of the product of 292 multiplied by c minus 4 and 7 is not 0 we know that the difference is not divisible with 50.
So the only answer I found is c = 7  and a = 2 so b = 9
But I can't prove it algebraically or write a better solution for this problem, I don't think testing is the best method for solving it...
 A: $$797a+50b=292c$$
Here we see that $a$ must be even, and if $a\ge4$ then the LHS is too big for the RHS to match in magnitude (remember that $0\le c\le 9$). Hence $a=2$:
$$1594+50b=292c$$
Now we see that $c$ must be at least $6$ so that the RHS can match the LHS in magnitude, and must be $2$ or $7$ to match in the last digit (the LHS's last digit must be $4$). Hence $c=7$, from which we get $b=9$ and the unique solution as $\overline{abc}=297$.
A: 
Not a 'real' answer, but it was too big for a comment.

I wrote and ran some Mathematica-code:
In[1]:=ParallelTable[
  If[TrueQ[8*(a*100 + b*10 + c*1) == 3*(100*c + 10*b + 1*a)], {a, b, 
    c}, Nothing], {a, 1, 9}, {b, 1, 9}, {c, 1, 9}] //. {} -> Nothing

Running the code gives:
Out[1]={{{{2, 9, 7}}}}

So, we can see that you're right!

When we expand this, with different values of the constants before the numbers we get:
In[2]:=ParallelTable[
  If[TrueQ[4*(a*100 + b*10 + c*1) == 7*(100*c + 10*b + 1*a)], {a, b, 
    c}, Nothing], {a, 1, 9}, {b, 1, 9}, {c, 1, 9}] //. {} -> Nothing

Out[2]={{{{2, 3, 1}}}, {{{4, 6, 2}}}, {{{6, 9, 3}}}}

