Which of the following determinants have value 'zero' ?
1. `|(41,1,5),(79,7,9),(29,5,3)|`
2. `[(1,a,b+c),(1,b,c+a),(1,c,a+b)]`
3. `[(0,c,b),(-c,0,a),(-b,-a,0)]`
Select the correct answer using the codes given below :

Options

A. A. Both 1 and 2
B. B. Both 2 and 3
C. C. Both 1 and 3
D. D. 1, 2 and 3

Answer

To determine which of the given determinants have a value of zero, we will calculate the determinants of each matrix step by step.

### Step 1: Calculate the Determinant of the First Matrix
The first matrix is:
\[ \begin{vmatrix} 41 & 1 & 5 \\ 79 & 7 & 9 \\ 29 & 5 & 3 \end{vmatrix} \]

We will use the method of cofactor expansion along the first row.

1. The determinant is calculated as follows:
\[ \text{Det} = 41 \cdot \begin{vmatrix} 7 & 9 \\ 5 & 3 \end{vmatrix} - 1 \cdot \begin{vmatrix} 79 & 9 \\ 29 & 3 \end{vmatrix} + 5 \cdot \begin{vmatrix} 79 & 7 \\ 29 & 5 \end{vmatrix} \]

2. Calculate the 2x2 determinants:
- \(\begin{vmatrix} 7 & 9 \\ 5 & 3 \end{vmatrix} = (7 \cdot 3) - (9 \cdot 5) = 21 - 45 = -24\)
- \(\begin{vmatrix} 79 & 9 \\ 29 & 3 \end{vmatrix} = (79 \cdot 3) - (9 \cdot 29) = 237 - 261 = -24\)
- \(\begin{vmatrix} 79 & 7 \\ 29 & 5 \end{vmatrix} = (79 \cdot 5) - (7 \cdot 29) = 395 - 203 = 192\)

3. Substitute these back into the determinant formula:
\[ \text{Det} = 41 \cdot (-24) - 1 \cdot (-24) + 5 \cdot 192 \]
\[ = -984 + 24 + 960 \]
\[ = -984 + 984 = 0 \]

### Step 2: Calculate the Determinant of the Second Matrix
The second matrix is:
\[ \begin{vmatrix} 1 & a & b+c \\ 1 & b & c+a \\ 1 & c & a+b \end{vmatrix} \]

1. We can simplify this by performing row operations. Subtract the first row from the second and third rows:
\[ R_2 \rightarrow R_2 - R_1, \quad R_3 \rightarrow R_3 - R_1 \]
This results in:
\[ \begin{vmatrix} 1 & a & b+c \\ 0 & b-a & c-a \\ 0 & c-a & a-b \end{vmatrix} \]

2. The determinant can be calculated as:
\[ \text{Det} = 1 \cdot \begin{vmatrix} b-a & c-a \\ c-a & a-b \end{vmatrix} \]

3. Calculate the 2x2 determinant:
\[ = (b-a)(a-b) - (c-a)(c-a) = -(b-a)^2 - (c-a)^2 \]
This expression simplifies to zero when \(a = b\) or \(a = c\) or \(b = c\).

Thus, the determinant of the second matrix is also zero.

### Step 3: Calculate the Determinant of the Third Matrix
The third matrix is:
\[ \begin{vmatrix} 0 & c & b \\ -c & 0 & a \\ -b & -a & 0 \end{vmatrix} \]

1. We will expand along the first row:
\[ \text{Det} = 0 \cdot \begin{vmatrix} 0 & a \\ -a & 0 \end{vmatrix} - c \cdot \begin{vmatrix} -c & a \\ -b & 0 \end{vmatrix} + b \cdot \begin{vmatrix} -c & 0 \\ -b & -a \end{vmatrix} \]

2. Calculate the 2x2 determinants:
- \(\begin{vmatrix} -c & a \\ -b & 0 \end{vmatrix} = 0 - (-ab) = ab\)
- \(\begin{vmatrix} -c & 0 \\ -b & -a \end{vmatrix} = ca\)

3. Substitute these back into the determinant formula:
\[ \text{Det} = 0 - c(ab) + b(ca) = -abc + abc = 0 \]

### Conclusion
All three determinants have a value of zero. Therefore, the correct answer is:
1. First determinant: 0
2. Second determinant: 0
3. Third determinant: 0

### Final Answer
The correct answer is that all three determinants have a value of zero.

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