If (i) A = 1, B = 0, C = 1, (ii) A = B = C = 1, (iii) A = B = C = 0 and (iv) A = 1 = B, C = 0 then which one of the following options will satisfy the expression, `X=bar(A.B.C)+bar(B.C.A)+bar(C.A.B)`

Options

A. (1, 0, 1, 1)
B. (1, 0, 1, 0)
C. (0, 1, 1, 1)
D. `(1, 1, 1, 0)

Explanation

To solve the expression \( X = \overline{A \cdot B \cdot C} + \overline{B \cdot C \cdot A} + \overline{C \cdot A \cdot B} \) for the given values of A, B, and C, we will evaluate the expression for each case step by step.

### Step 1: Evaluate for \( (i) A = 1, B = 0, C = 1 \)

1. Substitute the values into the expression:
\[ X = \overline{1 \cdot 0 \cdot 1} + \overline{0 \cdot 1 \cdot 1} + \overline{1 \cdot 1 \cdot 0} \]
2. Calculate each term:
- \( \overline{1 \cdot 0 \cdot 1} = \overline{0} = 1 \)
- \( \overline{0 \cdot 1 \cdot 1} = \overline{0} = 1 \)
- \( \overline{1 \cdot 1 \cdot 0} = \overline{0} = 1 \)
3. Combine the results:
\[ X = 1 + 1 + 1 = 1 \]

### Step 2: Evaluate for \( (ii) A = B = C = 1 \)

1. Substitute the values:
\[ X = \overline{1 \cdot 1 \cdot 1} + \overline{1 \cdot 1 \cdot 1} + \overline{1 \cdot 1 \cdot 1} \]
2. Calculate each term:
- \( \overline{1 \cdot 1 \cdot 1} = \overline{1} = 0 \)
- \( \overline{1 \cdot 1 \cdot 1} = \overline{1} = 0 \)
- \( \overline{1 \cdot 1 \cdot 1} = \overline{1} = 0 \)
3. Combine the results:
\[ X = 0 + 0 + 0 = 0 \]

### Step 3: Evaluate for \( (iii) A = B = C = 0 \)

1. Substitute the values:
\[ X = \overline{0 \cdot 0 \cdot 0} + \overline{0 \cdot 0 \cdot 0} + \overline{0 \cdot 0 \cdot 0} \]
2. Calculate each term:
- \( \overline{0 \cdot 0 \cdot 0} = \overline{0} = 1 \)
- \( \overline{0 \cdot 0 \cdot 0} = \overline{0} = 1 \)
- \( \overline{0 \cdot 0 \cdot 0} = \overline{0} = 1 \)
3. Combine the results:
\[ X = 1 + 1 + 1 = 1 \]

### Step 4: Evaluate for \( (iv) A = 1, B = 1, C = 0 \)

1. Substitute the values:
\[ X = \overline{1 \cdot 1 \cdot 0} + \overline{1 \cdot 0 \cdot 1} + \overline{0 \cdot 1 \cdot 1} \]
2. Calculate each term:
- \( \overline{1 \cdot 1 \cdot 0} = \overline{0} = 1 \)
- \( \overline{1 \cdot 0 \cdot 1} = \overline{0} = 1 \)
- \( \overline{0 \cdot 1 \cdot 1} = \overline{0} = 1 \)
3. Combine the results:
\[ X = 1 + 1 + 1 = 1 \]

### Summary of Results
- For \( (i) \): \( X = 1 \)
- For \( (ii) \): \( X = 0 \)
- For \( (iii) \): \( X = 1 \)
- For \( (iv) \): \( X = 1 \)

### Final Answer
The expression \( X \) satisfies the conditions for the inputs \( (i) \), \( (iii) \), and \( (iv) \) with \( X = 1 \) and for \( (ii) \) with \( X = 0 \).