Let `vec(a), vec(b) and vec(c)` be three mutually perpendicular vectors each of unit magnitud. If `vec(A)=vec(a)+vec(b)+vec(c), vec(B)=vec(a)-vec(b)+vec(c) and vec(C)=vec(a)-vec(b)-vec(c)`, then which one of the following is correct?

Options

A. `|vec(A)|gt|vec(B)|gt|vec(C)|`
B. `|vec(A)|=|vec(B)|!=|vec(C)|`
C. `|vec(A)|=|vec(B)|=|vec(C)|`
D. `|vec(A)|!=|vec(B)|!=|vec(C)|`

Explanation

To solve the problem, we need to determine the magnitudes of the vectors \(\vec{A}\), \(\vec{B}\), and \(\vec{C}\) given their definitions in terms of the mutually perpendicular unit vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\).

### Step-by-Step Solution:

1. **Define the vectors:**
\[ \vec{A} = \vec{a} + \vec{b} + \vec{c} \]
\[ \vec{B} = \vec{a} - \vec{b} + \vec{c} \]
\[ \vec{C} = \vec{a} - \vec{b} - \vec{c} \]

2. **Calculate the magnitude of \(\vec{A}\):**
\[ |\vec{A}| = |\vec{a} + \vec{b} + \vec{c}| \]
Since \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) are mutually perpendicular unit vectors, we can use the formula for the magnitude of the sum of vectors:
\[ |\vec{A}| = \sqrt{|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2} \]
Substituting the values:
\[ |\vec{A}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \]

3. **Calculate the magnitude of \(\vec{B}\):**
\[ |\vec{B}| = |\vec{a} - \vec{b} + \vec{c}| \]
Again, using the same formula:
\[ |\vec{B}| = \sqrt{|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2} = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3} \]

4. **Calculate the magnitude of \(\vec{C}\):**
\[ |\vec{C}| = |\vec{a} - \vec{b} - \vec{c}| \]
Using the same formula:
\[ |\vec{C}| = \sqrt{|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2} = \sqrt{1^2 + (-1)^2 + (-1)^2} = \sqrt{3} \]

5. **Conclusion:**
Since we have found that:
\[ |\vec{A}| = |\vec{B}| = |\vec{C}| = \sqrt{3} \]
Therefore, the correct relation is:
\[ \vec{A} = \vec{B} = \vec{C} \]

### Final Answer:
The correct option is that \(\vec{A} = \vec{B} = \vec{C}\).