<s>[INST] What can we do to solve the following equation with $6$ variables with some information provided? 
Q) There are unique integers $a_2, a_3, a_4, a_5, a_6, a_7$ such that $$\frac{a_2}{2!}+\frac{a_3}{3!}+\frac{a_4}{4!}+\frac{a_5}{5!}+\frac{a_6}{6!}+\frac{a_7}{7!}=\frac 57$$,where $0\le a_i < i$. Then the value of $a_2+a_3+a_4+a_5+a_6+a_7$(1)$8$(2)$9$(3)$10$(4)$11$

Answer: $(2)$
My workout
all these are unique integers hence all are having different values.
 now multiplying the big one equation by $7!$ 
$$7\left\{6!\frac{a_2}{2!}+6!\frac{a_3}{3!}+6!\frac{a_4}{4!}+6!\frac{a_5}{5!}+6!\frac{a_6}{6!}\right\}+a_7= 6!.5$$
which is like the remainder theorem which implies, $6!.5=3600$, when divided by 7 leaves remainder $a_7$ so dividing 3600 by 7 gives 2 hence $a_7=2$ . Now from inequality equation we can say that
$a_2\in\{0,1\}$
$a_3\in\{0,1,2\}$
$a_4\in\{0,1,2,3\}$
$a_5\in\{0,1,2,3,4\}$
$a_6\in\{0,1,2,3,4,5\}$
$a_7\in\{0,1,2,3,4,5,6\}$
so from all the data above i can conclude that $a_2=0, a_3=1, a_4=3, a_5=4, a_6=5, a_7=2$ hence summing them up $0+1+3+4+5+2=15$ which is none of the above option
 [/INST] 5xum has already pointed out your error.

Multiplying the both sides by $6!$ gives
$$\text{(integer)}+\frac{a_7}{7}=\frac{5\cdot 6!}{7}\quad\Rightarrow\quad a_7\equiv 5\cdot 6!\pmod 7\quad\Rightarrow\quad a_7=2$$
Multiplying the both sides by $7!/6$ gives
$$\text{(integer)}+\frac{7a_6+a_7}{6}=5!\times 5\quad\Rightarrow\quad 7a_6+a_7\equiv 0\pmod 6\quad\Rightarrow \quad a_6=4$$
Multiplying the both sides by $7!/5$ gives
$$\small\text{(integer)}+\frac{42a_5+7a_6+a_7}{5}=6\times 4!\times 5\quad\Rightarrow\quad 42a_5+7a_6+a_7\equiv 0\pmod 5\quad\Rightarrow\quad a_5=0$$
I think you can continue from here.
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