Statement I: If in a triangle `ABC, sin ^(2) A+sin ^(2)B+sin ^(2)C=2,` then one of the angles must be `90^(@).`
Statement II: In any triangle `ABC` `cos 2A+ cos 2B+cos 2C=-1-4 cos A cos B cos C`

Options

A. (a) Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I
B. (b) Both Statement I and Statement II are correct but Statement II is not the correct explanation of Statement I
C. (c) Statement I is correct but Statement II is incorrect
D. (d) Statement II is correct but Statement I is incorrect

Answer

To solve the problem, we need to analyze both statements and determine their validity step by step.

### Step 1: Analyze Statement II
We start with Statement II, which claims:
\[ \cos 2A + \cos 2B + \cos 2C = -1 - 4 \cos A \cos B \cos C \]

**Hint:** Recall that the sum of angles in a triangle is \( A + B + C = \pi \).

Using the identity for the cosine of a sum, we can express \( \cos 2A \) and \( \cos 2B \) in terms of \( A \) and \( B \):
\[ \cos 2A = 2 \cos^2 A - 1 \]
\[ \cos 2B = 2 \cos^2 B - 1 \]
\[ \cos 2C = 2 \cos^2 C - 1 \]

Substituting these into the left-hand side gives:
\[ (2 \cos^2 A - 1) + (2 \cos^2 B - 1) + (2 \cos^2 C - 1) \]
This simplifies to:
\[ 2(\cos^2 A + \cos^2 B + \cos^2 C) - 3 \]

**Hint:** Use the identity \( \cos^2 A + \cos^2 B + \cos^2 C = 1 + 2 \cos A \cos B \cos C \).

Now, we can substitute this identity into our equation:
\[ 2(1 + 2 \cos A \cos B \cos C) - 3 = -1 - 4 \cos A \cos B \cos C \]
This simplifies to:
\[ 2 + 4 \cos A \cos B \cos C - 3 = -1 - 4 \cos A \cos B \cos C \]
Thus, we have:
\[ -1 + 4 \cos A \cos B \cos C = -1 - 4 \cos A \cos B \cos C \]
Adding \( 4 \cos A \cos B \cos C \) to both sides gives:
\[ 0 = 0 \]
This confirms that Statement II is true.

### Step 2: Analyze Statement I
Now we move to Statement I, which claims:
\[ \sin^2 A + \sin^2 B + \sin^2 C = 2 \]

**Hint:** Use the identity \( \sin^2 A + \sin^2 B + \sin^2 C = 2 + 2 \cos A \cos B \cos C \).

From Statement II, we have:
\[ \sin^2 A + \sin^2 B + \sin^2 C = 2 + 2 \cos A \cos B \cos C \]

Setting this equal to 2 gives:
\[ 2 + 2 \cos A \cos B \cos C = 2 \]
This simplifies to:
\[ 2 \cos A \cos B \cos C = 0 \]

**Hint:** Consider the implications of the product being zero.

Since \( \cos A \cos B \cos C = 0 \), at least one of the angles \( A, B, \) or \( C \) must be \( 90^\circ \) (or \( \frac{\pi}{2} \) radians). This confirms that Statement I is also true.

### Conclusion
Both statements are correct, and Statement II correctly explains Statement I. Therefore, the correct answer is:
**Option A:** Both statements are correct, and Statement II is the correct explanation of Statement I.