Let the first term of the geometric progression be $a$ and the common ratio be $r$. Then, $a_1 = a$, $a_2 = ar$, $a_3 = ar^2$, $a_4 = ar^3$, $a_5 = ar^4$, and so on.

Given that $a_1 + a_3 + a_5 = 21$, we have $a + ar^2 + ar^4 = 21$, which can be written as $a(1 + r^2 + r^4) = 21$.

Also, given that $a_1 a_3 a_5 = 64$, we have $a(ar^2)(ar^4) = 64$, which simplifies to $a^3 r^6 = 64$. Taking the cube root of both sides, we get $ar^2 = 4$.

Now, we can substitute $ar^2 = 4$ into the first equation: $a(1 + r^2 + r^4) = 21$. Since $a = \frac{4}{r^2}$, we have $\frac{4}{r^2}(1 + r^2 + r^4) = 21$. Multiplying both sides by $r^2$, we get $4(1 + r^2 + r^4) = 21r^2$, which simplifies to $4 + 4r^2 + 4r^4 = 21r^2$. Rearranging the terms, we get $4r^4 - 17r^2 + 4 = 0$.

Let $x = r^2$. Then the equation becomes $4x^2 - 17x + 4 = 0$. We can factor this quadratic equation as $(4x - 1)(x - 4) = 0$. Thus, $x = \frac{1}{4}$ or $x = 4$. Since the geometric progression is increasing, $r > 1$, so $r^2 > 1$. Therefore, $r^2 = 4$, which means $r = 2$ (since $r>0$).

Now, we can find $a$ using $ar^2 = 4$. Since $r = 2$, we have $a(2^2) = 4$, so $4a = 4$, which means $a = 1$.

We want to find $a_1 + a_2 + a_3 = a + ar + ar^2 = 1 + 1(2) + 1(2^2) = 1 + 2 + 4 = 7$.