Why does the largest $x$ such that $a$, $b$ divided by $x$ leave the same remainder equal $a-b$? Suppose two numbers $a$ and $b$ as, $a=kq_1+r_1=3\times 17 + 1  = 52$ and $b = kq_2+r_2=3 \times 15 +1=46$.
It is clear that $52$ and $46$ leave the same reminder 1 when divided by $3$, because I designed them this way. But surprisingly however I design the numbers the largest $x$ which leaves the same reminder is $kq_1-kq_2=k(q_1-q_2)$. Why is that? In this case we have $52 = 6\times 8+ \color \red  4$ and $46 = 6\times 7 + \color \red 4$.
Now suppose there are three numbers $a$, $b$, $c$ and $x$(assuming $a>b>c \geq x$) such that $x$ leaves the same reminder when we divide each of $a,b$ and $c$ with it. $x$ is supposed to be the largest possible value that holds the assertion.  Now $x$ is given by the H.C.F of $a-b, a-c$ and $b-c$. Why is that? How can we prove this mathematically?

Your claim is not correct. Let $a=13$, $b=7$, $c=1$, $x=2$.
Clearly, $a=b=c = 1 \mod x$. So your assumptions are met.
Then
$$ a-b = 13-7 = 6$$
$$ b-c = 7-1 = 6 $$
$$ a-c = 13-1 = 12.$$
Obviously, $\mathrm{gcd}(6,6,12) = 6 \neq 2 = x$.
This leads to a contradiction, and hence your statement is not correct.
