Let `T_(n)` denotes the `n ^(th)` term of a G.P. with common ratio 2 and `(log _(2) (log _(3) ( log _(512) T _(100))))=1.` If three sides of a triangle ABC are the values of `(T _(1) + T _(2)), T_(2) and T_(3)` then area of the triangle is `(sqrt 2160)/(N),` where N is a positive integer. Find the remainder when N is divided by `2 ^(10).`

Answer

To solve the problem step by step, we will follow the same logical flow as in the video transcript, ensuring clarity and completeness.

### Step-by-Step Solution:

1. **Identify the General Term of the G.P.:**
Let \( T_n \) denote the \( n^{th} \) term of a geometric progression (G.P.) with a common ratio of 2. We can express the terms as:
\[ T_1 = a, \quad T_2 = 2a, \quad T_3 = 4a, \quad T_n = a \cdot 2^{n-1} \]

2. **Find \( T_{100} \):**
Using the formula for the \( n^{th} \) term:
\[ T_{100} = a \cdot 2^{99} \]

3. **Use the Given Logarithmic Equation:**
We are given:
\[ \log_2(\log_3(\log_{512}(T_{100}))) = 1 \]
Taking the antilogarithm of both sides:
\[ \log_3(\log_{512}(T_{100})) = 2 \]
Again taking the antilogarithm:
\[ \log_{512}(T_{100}) = 3^2 = 9 \]

4. **Convert the Logarithm Base:**
Since \( 512 = 2^9 \), we can rewrite the logarithm:
\[ \log_{512}(T_{100}) = \frac{\log_2(T_{100})}{\log_2(512)} = \frac{\log_2(T_{100})}{9} \]
Setting this equal to 9:
\[ \frac{\log_2(T_{100})}{9} = 9 \implies \log_2(T_{100}) = 81 \]

5. **Find \( T_{100} \):**
Taking the antilogarithm:
\[ T_{100} = 2^{81} \]
From our earlier expression for \( T_{100} \):
\[ a \cdot 2^{99} = 2^{81} \implies a = \frac{2^{81}}{2^{99}} = 2^{-18} \]

6. **Calculate \( T_1, T_2, T_3 \):**
Now we can find:
\[ T_1 = a = 2^{-18}, \quad T_2 = 2a = 2^{-17}, \quad T_3 = 4a = 2^{-16} \]

7. **Determine the Sides of the Triangle:**
The sides of triangle ABC are:
\[ T_1 + T_2 = 2^{-18} + 2^{-17} = \frac{1}{2^{18}} + \frac{2}{2^{18}} = \frac{3}{2^{18}} \]
Thus, the sides are:
\[ \text{Side 1} = \frac{3}{2^{18}}, \quad \text{Side 2} = \frac{2}{2^{18}}, \quad \text{Side 3} = \frac{4}{2^{18}} \]

8. **Calculate the Semi-perimeter \( S \):**
\[ S = \frac{\frac{3}{2^{18}} + \frac{2}{2^{18}} + \frac{4}{2^{18}}}{2} = \frac{9}{2^{19}} \]

9. **Apply Heron's Formula:**
The area \( \Delta \) of the triangle is given by:
\[ \Delta = \sqrt{S(S - a)(S - b)(S - c)} \]
Where \( a = \frac{3}{2^{18}}, b = \frac{2}{2^{18}}, c = \frac{4}{2^{18}} \):
\[ S - a = \frac{9}{2^{19}} - \frac{3}{2^{18}} = \frac{9 - 6}{2^{19}} = \frac{3}{2^{19}} \]
\[ S - b = \frac{9}{2^{19}} - \frac{2}{2^{18}} = \frac{9 - 4}{2^{19}} = \frac{5}{2^{19}} \]
\[ S - c = \frac{9}{2^{19}} - \frac{4}{2^{18}} = \frac{9 - 8}{2^{19}} = \frac{1}{2^{19}} \]

10. **Calculate the Area:**
\[ \Delta = \sqrt{\frac{9}{2^{19}} \cdot \frac{3}{2^{19}} \cdot \frac{5}{2^{19}} \cdot \frac{1}{2^{19}}} = \sqrt{\frac{135}{2^{76}}} = \frac{\sqrt{135}}{2^{38}} \]

11. **Relate to Given Area:**
We know:
\[ \Delta = \frac{\sqrt{2160}}{N} \]
Setting the two expressions for area equal:
\[ \frac{\sqrt{135}}{2^{38}} = \frac{\sqrt{2160}}{N} \]
Cross-multiplying gives:
\[ N \cdot \sqrt{135} = \sqrt{2160} \cdot 2^{38} \]

12. **Calculate \( N \):**
Solving for \( N \):
\[ N = \frac{\sqrt{2160} \cdot 2^{38}}{\sqrt{135}} \]
Simplifying \( \sqrt{2160} \) and \( \sqrt{135} \):
\[ \sqrt{2160} = \sqrt{216 \cdot 10} = \sqrt{216} \cdot \sqrt{10} = 6\sqrt{6} \cdot \sqrt{10} = 6\sqrt{60} = 6\cdot 2\sqrt{15} = 12\sqrt{15} \]
\[ \sqrt{135} = 3\sqrt{15} \]
Thus:
\[ N = \frac{12\sqrt{15} \cdot 2^{38}}{3\sqrt{15}} = 4 \cdot 2^{38} = 2^{40} \]

13. **Find the Remainder when \( N \) is Divided by \( 2^{10} \):**
\[ N = 2^{40} \implies \text{Remainder when } N \text{ is divided by } 2^{10} = 0 \]

### Final Answer:
The remainder when \( N \) is divided by \( 2^{10} \) is \( \boxed{0} \).