An approach to solving $2(2^{(a+b+c+2)} + 3\cdot 2^{(a+b+1)} + 9\cdot 2^a + 101)$ for multiples of 77

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In searching for counter examples to the Collatz Conjecture, I have come across a family of equations with the following form. $$77D = 2(2^{a+b+c+2} + 3 * 2^{a+b+1} + 9 * 2^a + 101)$$ Where $$0>a+b+c<=4$$and$$D>0$$Computationally solving for this example is relatively simple with $$a=1,b=2,c=1, D=6$$and$$a=0, b=1, c=2, D=4$$ However, as the equations of this form become larger and more useful, computation becomes impossible. Could anyone recommend an approach for determining if equations of this form have a solution? Or more ideally tell me such an approach does not exist, so that I can return to more fruitful endeavors.

Thank you for any time or insight you can provide.