Comprehension 2 Consider the system of linear equations `alphax+y+z=m x+alphay+z=n x+y+alphaz=p` If `alpha=1\ &\ m!=p` then the system of linear equations has- a. no solution b. `\ ` infinite solutions c. unique solution d. `\ ` unique solution `if\ p=n`

Options

A. The given system of equation has no solution if `alpha=-2` and `m+n+pne0`
B. The give system of equation has no solution if `alpha=1` and `m ne n` or `n ne p` or `p ne m`
C. The given system of equation has infinite solution if `alpha=-2` and `m+n+p=0`
D. The given system of equation has unique solution if `alpha=1` or `-2`

Explanation

To solve the given system of linear equations, we will first represent the equations in a standard form and then analyze the determinant to determine the number of solutions.

### Given System of Equations:
1. \( \alpha x + y + z = m \)
2. \( x + \alpha y + z = n \)
3. \( x + y + \alpha z = p \)

### Step 1: Substitute the value of \( \alpha \)
We are given that \( \alpha = 1 \). Substituting this into the equations, we get:
1. \( 1x + y + z = m \) → \( x + y + z = m \)
2. \( x + 1y + z = n \) → \( x + y + z = n \)
3. \( x + y + 1z = p \) → \( x + y + z = p \)

### Step 2: Analyze the equations
After substitution, we have:
1. \( x + y + z = m \)
2. \( x + y + z = n \)
3. \( x + y + z = p \)

### Step 3: Check for consistency
From the above equations, we see that:
- All three equations represent the same left-hand side \( x + y + z \).
- For the system to be consistent, the right-hand sides must also be equal.

### Step 4: Determine the number of solutions
Since we have:
- \( m \) must equal \( n \) and \( p \) for the system to have a unique solution.
- Given that \( m \neq p \), it implies that \( m \) and \( n \) cannot be equal either.

Thus, since \( m \neq p \), the system of equations is inconsistent.

### Conclusion
The system of linear equations has **no solution**.

### Final Answer:
**a. no solution**