Statement 1: If `a_(1)hati + a_(2)hatj + a_(3)hatk, vecbhati+b_(2)hatj + b_(3) hatk and c_(1)hati + c_(2)hatj + c_(3)hatk` are three mutually perpendicular unit vectors then `a_(1)hati + b_(1)hatj + c_(1)hatk,a_(2)hati +b_(2)hatj+c_(2) hatk and a_(3)hati + b_(3) hatj + c_(3) hatk` may be mutually perpendicular unit vectors.
Statement 2 : value of determinant and its transpose are the same.

Options

A. Both the statements are true and statement 2 is the correct explanation for statement 1.
B. Both statements are true but statement 2 is not the correct explanation for statement 1.
C. Statement 1 is true and Statement 2 is false
D. Statement 1 is false and Statement 2 is true.

Explanation

To analyze the given statements, we will break down the problem step by step.

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

These vectors are mutually perpendicular unit vectors. This means:
1. \( |\vec{a}| = |\vec{b}| = |\vec{c}| = 1 \)
2. \( \vec{a} \cdot \vec{b} = 0 \)
3. \( \vec{b} \cdot \vec{c} = 0 \)
4. \( \vec{c} \cdot \vec{a} = 0 \)

### Step 2: Forming New Vectors
Next, we form three new vectors:
- \( \vec{u_1} = a_1 \hat{i} + b_1 \hat{j} + c_1 \hat{k} \)
- \( \vec{u_2} = a_2 \hat{i} + b_2 \hat{j} + c_2 \hat{k} \)
- \( \vec{u_3} = a_3 \hat{i} + b_3 \hat{j} + c_3 \hat{k} \)

We need to determine if these new vectors \( \vec{u_1}, \vec{u_2}, \vec{u_3} \) can also be mutually perpendicular.

### Step 3: Checking Mutual Perpendicularity
To check if \( \vec{u_1}, \vec{u_2}, \vec{u_3} \) are mutually perpendicular, we compute their dot products:
1. \( \vec{u_1} \cdot \vec{u_2} \)
2. \( \vec{u_2} \cdot \vec{u_3} \)
3. \( \vec{u_3} \cdot \vec{u_1} \)

Calculating \( \vec{u_1} \cdot \vec{u_2} \):
\[ \vec{u_1} \cdot \vec{u_2} = (a_1 \hat{i} + b_1 \hat{j} + c_1 \hat{k}) \cdot (a_2 \hat{i} + b_2 \hat{j} + c_2 \hat{k}) = a_1 a_2 + b_1 b_2 + c_1 c_2 \]

For the vectors to be mutually perpendicular, this dot product must equal zero:
\[ a_1 a_2 + b_1 b_2 + c_1 c_2 = 0 \]

### Step 4: Determinant Condition
The condition for three vectors to be mutually perpendicular can also be represented using the determinant of a matrix formed by these vectors. The determinant of the matrix formed by the coefficients of the vectors must equal zero.

The determinant \( D \) of the original vectors is:
\[ D = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix} \]

### Step 5: Transpose Property
By the property of determinants, the value of a determinant remains the same when transposed. Therefore, if \( D = 0 \), then the determinant of the new vectors formed will also be zero.

### Conclusion
Since the determinant of the original vectors is zero, the new vectors formed from the components of the original vectors can also be mutually perpendicular. Thus, Statement 1 is correct.

### Statement 2
Statement 2 states that the value of a determinant and its transpose are the same. This is a well-known property of determinants and is indeed correct.

### Final Answer
Both Statement 1 and Statement 2 are correct.