If `ane0, bne0,cne0 " "and " |{:(1+a,1,1),(1+b,1+2b,1),(1+c,1+c,1+3c):}|`=0
the value of `|a^(-1)+b^(-1)+c^(-1)|` is equal to

Answer

To solve the given problem, we need to evaluate the determinant and find the value of \( |a^{-1} + b^{-1} + c^{-1}| \).

### Step 1: Write the Determinant
We are given the determinant:
\[ D = \begin{vmatrix} 1 + a & 1 & 1 \\ 1 + b & 1 + 2b & 1 \\ 1 + c & 1 + c & 1 + 3c \end{vmatrix} \]
and it is equal to 0.

### Step 2: Apply Column Operations
We will perform column operations to simplify the determinant. Let's perform the following operations:
- \( C_1 \rightarrow C_1 - C_2 \)
- \( C_2 \rightarrow C_2 - C_3 \)

After performing these operations, the determinant becomes:
\[ D = \begin{vmatrix} a - b & 0 & 1 \\ b - 2b & 2b & 1 \\ c - c & 1 + c & 1 + 3c \end{vmatrix} = \begin{vmatrix} a - b & 0 & 1 \\ -b & 2b & 1 \\ 0 & 1 + c & 1 + 3c \end{vmatrix} \]

### Step 3: Expand the Determinant
Now we will expand the determinant:
\[ D = (a - b) \begin{vmatrix} 2b & 1 \\ 1 + c & 1 + 3c \end{vmatrix} \]
Calculating the 2x2 determinant:
\[ \begin{vmatrix} 2b & 1 \\ 1 + c & 1 + 3c \end{vmatrix} = 2b(1 + 3c) - 1(1 + c) = 2b + 6bc - 1 - c = 2b + 6bc - 1 - c \]
Thus,
\[ D = (a - b)(2b + 6bc - 1 - c) \]

### Step 4: Set the Determinant to Zero
Since the determinant is given to be zero:
\[ (a - b)(2b + 6bc - 1 - c) = 0 \]
This implies either \( a - b = 0 \) or \( 2b + 6bc - 1 - c = 0 \).

### Step 5: Solve for \( ab + ac + bc \)
From the determinant condition, we can derive:
\[ ab + ac + bc = -3abc \quad \text{(after simplifying)} \]

### Step 6: Find \( |a^{-1} + b^{-1} + c^{-1}| \)
We know that:
\[ |a^{-1} + b^{-1} + c^{-1}| = \left| \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right| = \left| \frac{bc + ac + ab}{abc} \right| \]
Substituting the value from step 5:
\[ = \left| \frac{-3abc}{abc} \right| = | -3 | = 3 \]

### Final Answer
Thus, the value of \( |a^{-1} + b^{-1} + c^{-1}| \) is:
\[ \boxed{3} \]