Krdytkyu

$$ 144 cdot 12 - 75 cdot 23 = 3 $$
$$ 160 cdot 1 - 3 cdot 53 = 1 $$
$$ 160 cdot 1 - 53 ( 144 cdot 12 - 75 cdot 23 ) =1 $$
$$ 160 cdot 1 - 636 cdot 144 + 1219 cdot 75 = 1 $$

The shortest vector solution is
$$ a= 3, b= -2, c= -1, d= 1 $$
with
$$ 3 cdot 75 - 2 cdot 120 - 144 + 160 = 225 -240 -144 + 160 = 1 $$

The more interesting question is finding a basis for the lattice of integer vectors orthogonal to your vector. A basis is given by these three rows:

$$
left(
begin{array}{rrrr}
-175536 & 0 & 91585 & -144 \
-146280 & 1 & 76320 & -120 \
-91424 & 0 & 47700 & -75 \
end{array}
right)
$$
This three dimensional lattice has Gram determinant $66361$
Next is a reduced basis by the LLL algorithm.

$$
left(
begin{array}{rrrr}
0 & 4 & 0 & -3 \
0 & 2 & -5 & 3 \
8 & -1 & 0 & -3 \
end{array}
right)
$$

The Gram matrix for the reduced basis, still with determinant 66361, is

$$
left(
begin{array}{rrr}
25 & -1 & 5 \
-1 & 38 & -11 \
5 & -11 & 74 \
end{array}
right)
$$

There is a theorem involved here,
$$ 75^2 + 120^2 + 144^2 + 160^2 = 66361 $$

All solutions $(a,b,c,d)$ are given, with integer $s,t,u,$ by
$$ (3+8u,-2+4s+2t-u,-1-5t,1-3s+3t-3u) $$

First divide both sides by $3$ to get $$ 75a+120b+144c+160d=1$$

Now let $$a=b=c=x$$

Your equation changes to $$339x+160d=1$$

You can solve this one for $x$ and $d$ because $339$ and $160$ are relatively prime.
Back substitute and you have your solution.

You can systematically solve any such equation (or prove that there are no solutions) by the following:

Take any integers $a,b,c,d$. Then the following correspond:

• Solutions of $75a+120b+144c+160d = 1$
  


• Solutions of $120b+144c+160d equiv 1 pmod{75}$
  


• Solutions of $45b-6c+10d equiv 1 pmod{75}$
  


• Solutions of $45b-6c+10d+75p = 1$ where $p$ is an integer


• Solutions of $45b+10d+75p equiv 1 pmod{6}$
  


• Solutions of $3b+4d+3p equiv 1 pmod{6}$
  


• Solutions of $3b+4d+3p+6q = 1$ where $q$ is an integer


• Solutions of $4d equiv 1 pmod{3}$
  

And now you simply follow the reverse correspondences.

$color{#c00}6 = 2(75)!-!144,, $ $color{#0a0}{80} = 2(120)!-!160 $ so $,bbox[5px,border:1px solid red]{1 = 75!+!color{#c00}6!-!color{#0a0}{80} = 3(75)-2(120) -144 + 160}$

Remark $ $ Found by perusing coef remainders: $bmod 75: 144equiv -color{#c00}6,,$ $,bmod 120!: 160equiv -color{#0a0}{80}$

i.e. we applied a few (judicious) steps of the extended Euclidean algorithm.

Alternatively, applying the algorithm mechanically, reducing each argument by the least argument, we get a longer computation like that in user21820's answer, viz.

$$begin{align}
&gcd(color{#88f}{75},120,144,160) {rm so reducing} bmod color{#8af}{75}\
= &gcd (75, 45,,-color{#0a0}6, 10) {rm so reducing} bmod color{#0a0}{6} \
= &gcd( 3, 0, {-}color{#0a0}6, {-}color{#f4f}2) {rm so reducing} bmod color{#F4f}{2}\
= &gcd( color{#d00}{bf 1}, 0, 0, {-}2)\
end{align}qquadqquad $$

yielding a Bezout identity for $color{#d00}{bf 1}$ from the augmented matrix in the extended algorithm (follow the above link for a complete presentation with the augmented matrix displayed).

Here is a description of a program that I wrote to solve such problems. It is definitely not optimized.

I start by considering the equalities

begin{array}{r}
160 &= 160(1) &+& 144(0) &+& 120(0) &+& 75(0) \
144 &= 160(0) &+& 144(1) &+& 120(0) &+& 75(0) \
120 &= 160(0) &+& 144(0) &+& 120(1) &+& 75(0) \
75 &= 160(0) &+& 144(0) &+& 120(0) &+& 75(1) \
end{array}

This can be abstracted into the following partitioned array
begin{array}{r|rrrr}
160 & 1 & 0 & 0 & 0 \
144 & 0 & 1 & 0 & 0 \
120 & 0 & 0 & 1 & 0 \
75 & 0 & 0 & 0 & 1 \
end{array}

The important thing to remember is that, at any time, a row $fbox{n | d c b a}$ represents the equality $n = 160d + 144c + 120b + 75a$.

The "outer loop" of this algorithm assumes that the left column is in descending order.

The first step is to reduce the three upper rows so that the entries in the first column are all less than the bottom left number, $75$. For the first row, we know that
$160 = 2 times 75 + 10$ so we replace row $1$ ($R1$) with $R1 - 2R4$, getting $fbox{10 | 1 0 0 -2}$. Negative remainders are allowed if their absolute value is less than the positive remainder. So since $144 = 75(2)-6$, we replace the second row with $R2 - 2R4$, getting $fbox{-6 | 0 1 0 -2}$. Similarly the third row becomes $R3 - 24$, which is $fbox{-30 | 0 0 1 -2}$. So we now have

begin{array}{r|rrrr}
10 & 1 & 0 & 0 & -2 \
-6 & 0 & 1 & 0 & -2 \
-30 & 0 & 0 & 1 & -2 \
75 & 0 & 0 & 0 & 1 \
end{array}

Next, we make the substitution $Rk to -Rk$ for any row with a negative first element and we then sort the array in decreasing order of the first element. We end up with

begin{array}{r|rrrr}
75 & 0 & 0 & 0 & 1 \
30 & 0 & 0 & -1 & 2 \
10 & 1 & 0 & 0 & -2 \
6 & 0 & -1 & 0 & 2 \
end{array}

After the next pass through the loop, we get

begin{array}{r|rrrr}
3 & 0 & 12 & 0 & -23 \
0 & 0 & 5 & -1 & -8 \
-2 & 1 & 2 & 0 & -6 \
6 & 0 & -1 & 0 & 2 \
end{array}

which "sorts" to

begin{array}{r|rrrr}
6 & 0 & -1 & 0 & 2 \
3 & 0 & 12 & 0 & -23 \
2 & -1 & -2 & 0 & 6 \
0 & 0 & 5 & -1 & -8 \
end{array}

The next outer loop will proceed as before except that we will make our adjustments with respect to the third row instead of the fourth row. we finally end up with

begin{array}{r|rrrr}
1 & 1 & 14 & 0 & -29 \
0 & -3 & -30 & 0 & 64 \
0 & 3 & 5 & 0 & -6 \
0 & 0 & 5 & -1 & -8 \
end{array}

The tells is that the general solution is (if I haven't messed up my math)

$(d,c,b,a) = (1, 14 , 0 , -29) + u(-3, -30, 0, 64) + v(3, 5, 0, -6) + w(0, 5, -1, -8)$