Step 1: Let the first three terms of the GP be a, ar, $ar^{2}$, where 'a' is the first term and 'r' is the common ratio.

Step 2: Given the sum of the first three terms is 26: a + ar + $ar^{2}$ = 26 => a(1 + r + $r^{2}$) = 26 ... (1).

Step 3: Given the sum of their squares is 364: $a^{2}$ + $a^{2r}$^2 + $a^{2r}$^4 = 364 => $a^{2}$(1 + $r^{2}$ + $r^{4}$) = 364 ... (2).

Step 4: Divide equation (2) by equation (1): $a^{2}$(1 + $r^{2}$ + $r^{4}$) / a(1 + r + $r^{2}$) = 364 / 26. Using the identity (1 + $r^{2}$ + $r^{4}$) / (1 + r + $r^{2}$) = 1 - r + $r^{2}$, we get a(1 - r + $r^{2}$) = 14 ... (3).

Step 5: Subtract equation (3) from equation (1): a(1 + r + $r^{2}$) - a(1 - r + $r^{2}$) = 26 - 14, which simplifies to 2ar = 12. Thus, the second term ar = 6.

Step 6: Substitute ar = 6 into equation (1): a + 6 + $ar^{2}$ = 26 => a + $ar^{2}$ = 20. Since a = 6/r and $ar^{2}$ = 6r, we get 6/r + 6r = 20.

Step 7: Multiply by r and simplify: 6 + $6r^{2}$ = 20r => $3r^{2}$ - 10r + 3 = 0. Factoring gives (3r - 1)(r - 3) = 0, so r = 3 or r = 1/3.

Step 8: If r = 3, a = 2, and the terms are 2, 6, 18. If r = 1/3, a = 18, and the terms are 18, 6, 2. In both cases, the terms of the GP are 2, 6, and 18.