Statement-I `a_1hat(i)+a_2hat(j)+a_3hat(k), b_1hat(i)+b_2hat(j)+b_3hat(k) and c_1hat(i)+c_2hat(j)+c_3hat(k)` are three mutually perpendicular unit vector, then `a_1hat(i)+b_1hat(j)+c_1hat(k), a_2hat(i)+b_2hat(j)+c_2hat(k) and a_3hat(i)+b_3hat(j)+3hat(k)` may be mutually perpendicular unit vectors.
Statement-II Value of determinant and its transpose are the same .

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 given problem, we need to analyze the statements and determine if they are correct.

### Step 1: Understanding the Vectors
We have three unit vectors:
- \( \mathbf{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k} \)
- \( \mathbf{b} = b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k} \)
- \( \mathbf{c} = c_1 \hat{i} + c_2 \hat{j} + c_3 \hat{k} \)

These vectors are mutually perpendicular unit vectors. This means:
- \( \|\mathbf{a}\| = \|\mathbf{b}\| = \|\mathbf{c}\| = 1 \)
- \( \mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{c} = \mathbf{c} \cdot \mathbf{a} = 0 \)

### Step 2: Forming New Vectors
Next, we need to check the following vectors:
1. \( \mathbf{u} = a_1 \hat{i} + b_1 \hat{j} + c_1 \hat{k} \)
2. \( \mathbf{v} = a_2 \hat{i} + b_2 \hat{j} + c_2 \hat{k} \)
3. \( \mathbf{w} = a_3 \hat{i} + b_3 \hat{j} + c_3 \hat{k} \)

We need to determine if these vectors can also be mutually perpendicular.

### Step 3: Checking Mutual Perpendicularity
To check if \( \mathbf{u}, \mathbf{v}, \mathbf{w} \) are mutually perpendicular, we need to compute their dot products:
- \( \mathbf{u} \cdot \mathbf{v} \)
- \( \mathbf{v} \cdot \mathbf{w} \)
- \( \mathbf{w} \cdot \mathbf{u} \)

Calculating these:
1. \( \mathbf{u} \cdot \mathbf{v} = (a_1 \hat{i} + b_1 \hat{j} + c_1 \hat{k}) \cdot (a_2 \hat{i} + b_2 \hat{j} + c_2 \hat{k}) = a_1 a_2 + b_1 b_2 + c_1 c_2 \)
2. \( \mathbf{v} \cdot \mathbf{w} = (a_2 \hat{i} + b_2 \hat{j} + c_2 \hat{k}) \cdot (a_3 \hat{i} + b_3 \hat{j} + c_3 \hat{k}) = a_2 a_3 + b_2 b_3 + c_2 c_3 \)
3. \( \mathbf{w} \cdot \mathbf{u} = (a_3 \hat{i} + b_3 \hat{j} + c_3 \hat{k}) \cdot (a_1 \hat{i} + b_1 \hat{j} + c_1 \hat{k}) = a_3 a_1 + b_3 b_1 + c_3 c_1 \)

For \( \mathbf{u}, \mathbf{v}, \mathbf{w} \) to be mutually perpendicular, all these dot products must equal zero:
- \( a_1 a_2 + b_1 b_2 + c_1 c_2 = 0 \)
- \( a_2 a_3 + b_2 b_3 + c_2 c_3 = 0 \)
- \( a_3 a_1 + b_3 b_1 + c_3 c_1 = 0 \)

### Step 4: Determinant and Its Transpose
The second statement mentions that the value of a determinant and its transpose are the same. This is a known property of determinants, which states that for any square matrix \( A \), \( \text{det}(A) = \text{det}(A^T) \).

### Conclusion
Based on the analysis:
- Statement I is correct if the conditions for mutual perpendicularity of the new vectors are satisfied.
- Statement II is a true mathematical property.

Thus, both statements are correct, and Statement II provides a correct explanation for Statement I.

### Final Answer
Both Statement I and Statement II are correct, and Statement II is a correct explanation of Statement I.

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