I reduced the equation:
$84x \equiv 68\mod 400$
$21x \equiv 17\mod 100$
$100 = 4 \times 21 + 16$
$21 = 1 \times 16 + 5$
$16 = 3 \times 5 + 1$
so we have...
$16 = 100 - 4 \times 21$
$5 = 21 - 1 \times 16$
$1 = 16 - 3 \times 5$
Now, I need the equation in the form of $21v + 100w = 1$ and I have to find v and w
$1=16-3 \times 5$
$=16-3(21-1 \times 16)$
$=16 - 3 \times 21 +3 \times 16$
$=100 - 4 \times 21 -3 \times 21 +3 \times 16$
At this point I don't know what to do. Please help.
The congruence $84x \equiv 68 \pmod{400}$ means $84x = 68 + 400t$ for some integer $t$. Dividing both sides of the equation by $4$ yields $21x = 17 + 100t$, so your conclusion that $21x \equiv 17 \pmod{100}$ is valid, if not fully justified. The reason it is useful to write the congruence in the form $21x \equiv 17 \pmod{100}$ is that $\gcd(21, 100) = 1$, so $21$ has a multiplicative inverse modulo $100$ and the congruence $21x \equiv 17 \pmod{100}$ has a unique solution. We can find that solution by finding the multiplicative inverse of $21 \pmod{100}$. To do so, we use the extended Euclidean algorithm.
You used the extended Euclidean algorithm to correctly obtain $$1 = 16 - 3 \cdot 21 + 3 \cdot 16$$ which we can simplify as $$1 = 4 \cdot 16 - 3 \cdot 21$$ Since $16 = 100 - 4 \cdot 21$, we obtain \begin{align*} 1 & = 4(100 - 4 \cdot 21) - 3 \cdot 21\\ & = 4 \cdot 100 - 16 \cdot 21 - 3 \cdot 21\\ & = 4 \cdot 100 - 19 \cdot 21 \end{align*} Thus, $-19$ is the multiplicative inverse of $21 \pmod{100}$. Multiplying both sides of the congruence $21x \equiv 17 \pmod{100}$ by $-19$ yields $$x \equiv -323 \equiv -323 + 4 \cdot 100 \equiv 77 \pmod{100}$$ However, we want the solutions of the congruence $84x \equiv 68 \pmod{400}$. Thus, we need all values of $x \in \{0, 1, 2, 3, \ldots, 399\}$ that satisfy $x \equiv 77 \pmod{100}$. Since $x \equiv 77 \pmod{100}$ means there exists an integer $k$ such that $x = 77 + 100k$. If we take $k = 0, 1, 2, 3$, respectively, we obtain the solutions
\begin{align*} x & \equiv 77 \pmod{400}\\ & \equiv 177 \pmod{400}\\ & \equiv 277 \pmod{400}\\ & \equiv 377 \pmod{400} \end{align*} of the congruence $84x \equiv 68 \pmod{400}$. Notice that there are no more solutions since any other number that satisfies $x \equiv 77 \pmod{100}$ is equivalent to one of these four values modulo $400$. For instance, if we take $k = 4$, we obtain $77 + 4 \cdot 100 \equiv 77 + 400 \equiv 77 \pmod{400}$.
Check: $84 \cdot 77 \equiv 6468 \equiv 16 \cdot 400 + 68 \equiv 68 \pmod{400}$. The other solutions can be checked similarly by direct substitution.