Rumination: Skip to the solution if you do not want to know what led me to the solution. At first sight, I was also intimidated by the problem only to realise that there is a pattern. That is simply because, I did not want to solve this problem in a way that would make me question myself. These are powers of natural numbers. And once you see this, the solution is almost immediate.
Solution: Clearly, the equations are as follows.
$$ a r^0 - br^1 + cr^2 -dr^3+er^4 = r^7 $$
for $r=1,2,3,4,5.$ In other words, the zeros of the equation
$$ r^7- er^4 + dr^3 - cr^2 + br -a = 0 $$
We know that this is a seven degree polynomial whose leading coefficient is $1$ and whose five of the zeros are $1,2,3,4,5.$ Let the other zeros be $\alpha, \beta$
$$ \implies r^7- er^4 + dr^3 - cr^2 + br -a = (r-1)(r-2)(r-3)(r-4)(r-5)(r-\alpha)(r-\beta) $$
Since the coefficients of $r^6$ and $r^5$ are $0,$ we compare the coefficients to get
$$ 1+2+3+4+5+\alpha + \beta = 0 \implies \alpha + \beta = -15, \ \text{and} $$
$$ \alpha \cdot \beta = 140 $$
Put $r=0$ in the equation to get
$$ - a = (-1)(-1)(-3)(-4)(-5)(\alpha \cdot \beta) $$
$$ \implies a= 140 \cdot 5! = 16800. $$
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