We can get a solution with three applications of the Euclidean Algorithm, I don't know if it possible to do any better.

To begin with we may as well divide throughout by 3,

75a + 120b +144c + 160d = 1.

Next, group as

3(25a + 48c) + 40(3b + 4d) = 1.

There are probably other groupings available, you just have to see that the three pairs of numbers (3,40), (25,48), (3,4) are each relativley prime.

Now rewrite as 3A + 40B = 1 and use the EA to find values for A and B.

A = -13, B = 1 are solutions (and there will be an infinity of others).

That gets you

25a + 48c = -13, and 3b + 4d = 1.

Solve these to get (again with an infinity of other possibles) a = 299, b = -1, c = -156 and d = 1.