Statement 1: `veca, vecb and vecc` arwe three mutually perpendicular unit vectors and `vecd` is a vector such that `veca, vecb, vecc and vecd` are non- coplanar. If `[vecd vecb vecc] = [vecdvecavecb] = [vecdvecc veca] = 1, " then " vecd= veca+vecb+vecc`
Statement 2: `[vecd vecb vecc] = [vecd veca vecb] = [vecdveccveca] Rightarrow vecd` is equally inclined to `veca, vecb and vecc`.

Options

A. (a) Both the statements are true and statement 2 is the correct explanation for statement 1.
B. (b) Both statements are true but statement 2 is not the correct explanation for statement 1.
C. (c) Statement 1 is true and Statement 2 is false
D. (d) Statement 1 is false and Statement 2 is true.

Explanation

To solve the problem, we need to analyze the given statements about the vectors \(\vec{a}\), \(\vec{b}\), \(\vec{c}\), and \(\vec{d}\).

### Step 1: Understanding the Given Information
We know:
- \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) are mutually perpendicular unit vectors.
- \(\vec{d}\) is a vector such that \(\vec{a}\), \(\vec{b}\), \(\vec{c}\), and \(\vec{d}\) are non-coplanar.
- The scalar triple products are given as:
\[ [\vec{d}, \vec{b}, \vec{c}] = [\vec{d}, \vec{a}, \vec{b}] = [\vec{d}, \vec{c}, \vec{a}] = 1 \]

### Step 2: Analyzing Statement 1
We need to show that if the above conditions hold, then \(\vec{d} = \vec{a} + \vec{b} + \vec{c}\).

1. **Assume \(\vec{d}\) in terms of \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\)**:
\[ \vec{d} = \lambda_1 \vec{a} + \lambda_2 \vec{b} + \lambda_3 \vec{c} \]

2. **Calculate the scalar triple product \([\vec{d}, \vec{b}, \vec{c}]\)**:
\[ [\vec{d}, \vec{b}, \vec{c}] = \vec{d} \cdot (\vec{b} \times \vec{c}) \]
Since \(\vec{b}\) and \(\vec{c}\) are perpendicular unit vectors, \(\vec{b} \times \vec{c} = \vec{a}\) and its magnitude is 1.

Thus,
\[ [\vec{d}, \vec{b}, \vec{c}] = \vec{d} \cdot \vec{a} = \lambda_1 \]
Given that this equals 1, we have:
\[ \lambda_1 = 1 \]

3. **Calculate the scalar triple product \([\vec{d}, \vec{a}, \vec{b}]\)**:
\[ [\vec{d}, \vec{a}, \vec{b}] = \vec{d} \cdot (\vec{a} \times \vec{b}) = \vec{d} \cdot \vec{c} = \lambda_3 \]
Given that this equals 1, we have:
\[ \lambda_3 = 1 \]

4. **Calculate the scalar triple product \([\vec{d}, \vec{c}, \vec{a}]\)**:
\[ [\vec{d}, \vec{c}, \vec{a}] = \vec{d} \cdot (\vec{c} \times \vec{a}) = \vec{d} \cdot \vec{b} = \lambda_2 \]
Given that this equals 1, we have:
\[ \lambda_2 = 1 \]

5. **Conclusion for Statement 1**:
Therefore, we have:
\[ \vec{d} = 1 \cdot \vec{a} + 1 \cdot \vec{b} + 1 \cdot \vec{c} = \vec{a} + \vec{b} + \vec{c} \]
Hence, Statement 1 is **true**.

### Step 3: Analyzing Statement 2
We need to determine if \([\vec{d}, \vec{b}, \vec{c}] = [\vec{d}, \vec{a}, \vec{b}] = [\vec{d}, \vec{c}, \vec{a}]\) implies that \(\vec{d}\) is equally inclined to \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\).

1. **Understanding Equal Inclination**:
If \(\vec{d}\) is equally inclined to \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\), it means the angles between \(\vec{d}\) and each of the unit vectors are the same.

2. **Using the Scalar Triple Product**:
Since all three scalar triple products are equal to 1, it indicates that the projection of \(\vec{d}\) onto the axes defined by \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) are equal, which implies equal inclination.

3. **Conclusion for Statement 2**:
Thus, Statement 2 is also **true**.

### Final Conclusion
Both Statement 1 and Statement 2 are true, and Statement 2 is indeed the correct explanation of Statement 1.