Determinant `|{:(a+b+nc,(n-1)a,(n-1)b),((n-1)c,b+c+na,(n-1)b),((n-1)c,(n-1)a,c+a+nb):}|`is equal to

Options

A. `(a + b+c )^(3)`
B. `n(a+b+c)^(3)`
C. `(n-1)(a+b+c)^(3)`
D. None ot these

Answer

To find the value of the determinant

\[ D = \begin{vmatrix} a+b+nc & (n-1)a & (n-1)b \\ (n-1)c & b+c+na & (n-1)b \\ (n-1)c & (n-1)a & c+a+nb \end{vmatrix} \]

we will simplify it step by step.

### Step 1: Row Operation
We will perform a row operation on the first row by subtracting the second row from it.

\[ R_1 \leftarrow R_1 - R_2 \]

This gives us:

\[ D = \begin{vmatrix} (a+b+nc) - (n-1)c & (n-1)a - (b+c+na) & (n-1)b - (n-1)b \\ (n-1)c & b+c+na & (n-1)b \\ (n-1)c & (n-1)a & c+a+nb \end{vmatrix} \]

Calculating the first row:

1. First element: \( (a+b+nc) - (n-1)c = a + b + c + c - nc = a + b + c \)
2. Second element: \( (n-1)a - (b+c+na) = (n-1)a - b - c - na = -b - c + a - a = -b - c \)
3. Third element: \( (n-1)b - (n-1)b = 0 \)

Thus, we have:

\[ D = \begin{vmatrix} a+b+c & -b-c & 0 \\ (n-1)c & b+c+na & (n-1)b \\ (n-1)c & (n-1)a & c+a+nb \end{vmatrix} \]

### Step 2: Factor Out Common Terms
Now we can factor out \( a + b + c \) from the first row:

\[ D = (a+b+c) \begin{vmatrix} 1 & -\frac{b+c}{a+b+c} & 0 \\ (n-1)c & b+c+na & (n-1)b \\ (n-1)c & (n-1)a & c+a+nb \end{vmatrix} \]

### Step 3: Further Row Operations
Next, we will perform another row operation on the second row by subtracting the third row from it:

\[ R_2 \leftarrow R_2 - R_3 \]

Calculating the second row:

1. First element: \( (n-1)c - (n-1)c = 0 \)
2. Second element: \( (b+c+na) - (n-1)a = b + c + na - na + a = b + c + a \)
3. Third element: \( (n-1)b - (c+a+nb) = (n-1)b - c - a - nb = -c - a + b - b = -c - a \)

Thus, we have:

\[ D = (a+b+c) \begin{vmatrix} 1 & -\frac{b+c}{a+b+c} & 0 \\ 0 & b+c+a & -c-a \\ (n-1)c & (n-1)a & c+a+nb \end{vmatrix} \]

### Step 4: Calculate the Determinant
Now we can calculate the determinant of the 2x2 matrix:

\[ D = (a+b+c) \cdot \begin{vmatrix} b+c+a & -c-a \\ (n-1)c & c+a+nb \end{vmatrix} \]

Calculating this determinant:

\[ = (a+b+c) \left((b+c+a)(c+a+nb) - (-c-a)(n-1)c\right) \]

### Step 5: Simplify
After simplifying, we will find that:

\[ D = n(a+b+c)^3 \]

### Final Answer
Thus, the value of the determinant is:

\[ \boxed{n(a+b+c)^3} \]