`|[1/c,1/c,-(a+b)/c^2],[-(b+c)/a^2,1/a,1/a],[(-b(b+c))/(a^2c),(a+2b+c)/(ac),(-b(a+b))/(ac^2)]|` is

Options

A. (a) dependent on `a,b,c`
B. (b) dependent on `a`
C. (c) dependent on `b`
D. (d) independent on `a,b` and `c`

Explanation

To solve the given determinant

\[ D = \begin{vmatrix} \frac{1}{c} & \frac{1}{c} & -\frac{(a+b)}{c^2} \\ -\frac{(b+c)}{a^2} & \frac{1}{a} & \frac{1}{a} \\ -\frac{b(b+c)}{(a^2c)} & \frac{(a+2b+c)}{(ac)} & -\frac{b(a+b)}{(ac^2)} \end{vmatrix} \]

we will simplify it step by step.

### Step 1: Multiply Columns by Constants
We will multiply the first column by \( A \), the second column by \( B \), and the third column by \( C \). This gives us:

\[ D = \frac{1}{ABC} \begin{vmatrix} \frac{A}{c} & \frac{B}{c} & -\frac{(a+b)C}{c^2} \\ -\frac{(b+c)A}{a^2} & \frac{B}{a} & \frac{B}{a} \\ -\frac{b(b+c)A}{(a^2c)} & \frac{(a+2b+c)B}{(ac)} & -\frac{b(a+b)C}{(ac^2)} \end{vmatrix} \]

### Step 2: Simplify the Determinant
Now we can simplify the determinant. The first column becomes:

\[ \begin{vmatrix} \frac{A}{c} & \frac{B}{c} & -\frac{(a+b)C}{c^2} \\ -\frac{(b+c)A}{a^2} & \frac{B}{a} & \frac{B}{a} \\ -\frac{b(b+c)A}{(a^2c)} & \frac{(a+2b+c)B}{(ac)} & -\frac{b(a+b)C}{(ac^2)} \end{vmatrix} \]

### Step 3: Perform Column Operations
Next, we will perform column operations. We can replace column 1 with the sum of columns 1, 2, and 3:

\[ C_1 \rightarrow C_1 + C_2 + C_3 \]

This gives us:

\[ D = \frac{1}{ABC} \begin{vmatrix} 0 & \frac{B}{c} & -\frac{(a+b)C}{c^2} \\ 0 & \frac{B}{a} & \frac{B}{a} \\ 0 & \frac{(a+2b+c)B}{(ac)} & -\frac{b(a+b)C}{(ac^2)} \end{vmatrix} \]

### Step 4: Evaluate the Determinant
Since the first column is now all zeros, the determinant evaluates to zero:

\[ D = 0 \]

### Conclusion
Thus, the determinant is independent of \( A \), \( B \), and \( C \).

### Final Answer
The value of the determinant is:

\[ \boxed{0} \]