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?

Here's my attempt at a step-by-step answer that does not use the notion of modular arithmetic.

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 $46 = 6\times 7 + \color \red 4$.

Let $a$ and $x$ be positive integers.  The integer quotient of $a$ and $x$ is the largest integer $q$ such that $qx\le a$.  The remainder of $a$ on division by $x$ is $r=a-qx$; it necessarily satisfies $0\le r<x$.
Observe first that $kq_1-kq_2=(a-r)-(b-r)=a-b$, where $r$ is the common remainder of $a$ and $b$ on division by $k$.  So your statement could be rephrased as "the largest $x$ that leaves the same remainder is $a-b$."  We prove this.
Let $a>b$, and $x$ be positive integers.  Let $q_a$ and $q_b$ be the corresponding integer quotients and let $r_a$ and $r_b$ be the corresponding remainders on division by $x$.  
For $a$ and $b$ to have the same remainder on division by $x$ means $r_a=r_b$ which means that $a-q_ax=b-q_bx$.  This implies that $a-b=x(q_a-q_b)$.  So if $a$ and $b$ have the same remainder on division by $x$, then $a-b$ is an integer multiple of $x$.
Conversely, if $a-b$ is an integer multiple of $x$, then $a-b=qx$ and therefore $a=b+qx$ for some integer $q$.  Since $r_a=a-q_ax$ and $r_b=b-q_bx$, we have
$$
r_a=b+qx-q_ax=r_b+q_bx+qx-q_ax=r_b+(q_b+q-q_a)x.
$$
We claim that $q_b+q-q_a=0$, and therefore $r_a=r_b$.  This follows since $0\le r_a,r_b<x$, so $-x<r_a-r_b<x$.  The only integer multiple of $x$ in this range is $0\cdot x$.  We have shown that if $a-b$ is a multiple of $x$, then $r_a=r_b$.
So $a$ and $b$ have the same remainder on division by $x$ if and only if $a-b$ is divisible by $x$.  The largest $x$ such that $a$ and $b$ have the same remainder on division by $x$ must therefore be the largest integer that divides $a-b$.  This is $a-b$ itself.

Now suppose there are three numbers $a$, $b$, $c$ and $x$(assuming $a>b>c>x$) such that $x$ leaves the same reminder when we divide each of $a,b$ and $c$ with it. 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?

We build on the previous proof to prove this one.  Any two of $a$, $b$, $c$ have the same remainder on division by $x$ if and only if their difference is divisible by $x$.  So $a$, $b$, and $c$ all have the same remainder on division by $x$ if and only if all of the differences, $a-b$, $a-c$, and $b-c$ are divisible by $x$.  The largest $x$ such that $a$, $b$, and $c$ all have the same remainder on division by $x$ is therefore the largest $x$ such that $x$ divides all three of $a-b$, $a-c$, and $b-c$.  By definition of greatest common factor, this is the greatest common factor of $a-b$, $a-c$, and $b-c$.
