If `|(a^2+1,ab,ac),(ab,b^2+1,bc),(ac,bc,c^2+1)|=1` , where a,b,c are real , then

Options

A. `a+b+c = 0 `
B. `a+b+c = 1`
C. `a^2+b^2+c^2=0`
D. `a=b=c=0`

Answer

To solve the determinant equation given by

\[ |(a^2+1, ab, ac), (ab, b^2+1, bc), (ac, bc, c^2+1)| = 1, \]

we will follow these steps:

### Step 1: Multiply Rows by Variables
Multiply the first row by \(a\), the second row by \(b\), and the third row by \(c\):

\[ | (a(a^2+1), a(ab), a(ac)), (b(ab), b(b^2+1), b(bc)), (c(ac), c(bc), c(c^2+1)) |. \]

This results in:

\[ | (a^3 + a, a^2b, a^2c), (ab^2, b^3 + b, b^2c), (ac^2, bc^2, c^3 + c) |. \]

### Step 2: Apply Row Operations
Now, we will perform the row operation \(R_1 \rightarrow R_1 + R_2 + R_3\):

The first row becomes:

\[ R_1 = (a^3 + a + ab^2 + b^3 + b + ac^2 + bc^2 + c^3 + c). \]

### Step 3: Factor Out Common Terms
We can factor out \(1 + a^2 + b^2 + c^2\) from the first row:

\[ |(1 + a^2 + b^2 + c^2, a^2b, a^2c), (ab^2, b^3 + b, b^2c), (ac^2, bc^2, c^3 + c)|. \]

### Step 4: Factor Out Columns
Next, we can factor out \(a\), \(b\), and \(c\) from the respective columns:

\[ = abc |(1 + a^2 + b^2 + c^2, b, c), (b, b^2 + 1, c), (c, b, c^2 + 1)|. \]

### Step 5: Evaluate the Determinant
Now, we need to evaluate the determinant:

\[ |(1 + a^2 + b^2 + c^2, b, c), (b, b^2 + 1, c), (c, b, c^2 + 1)|. \]

### Step 6: Set the Determinant Equal to 1
We know that the determinant equals 1:

\[ abc(1 + a^2 + b^2 + c^2) = 1. \]

### Step 7: Solve for \(a^2 + b^2 + c^2\)
From the equation above, we can isolate \(1 + a^2 + b^2 + c^2\):

\[ 1 + a^2 + b^2 + c^2 = \frac{1}{abc}. \]

### Step 8: Analyze the Result
For \(a^2 + b^2 + c^2\) to equal zero, it must be that \(a = b = c = 0\) since squares of real numbers are non-negative. Thus, we conclude:

\[ a^2 + b^2 + c^2 = 0 \implies a = 0, b = 0, c = 0. \]

### Final Answer
The correct options are:

1. \(A^2 + B^2 + C^2 = 0\) (True)
2. \(A = B = C = 0\) (True)