If `Delta_n=|{:(a^(2)+n,ab,ac),(ab,b^(2)+n,bc),(ac,bc,c^(2)+n):}|,n in N` and the equation
`x^(3)-lambdax^(2)+11x-6=0` has roots `a,b,c` where `a,b,c` are in AP.
The value of `sum_(r=1)^(30)((27Delta_(r)-Delta_(3r))/(27r^(2))) ` is

Options

A. 130
B. 190
C. 280
D. 340

Answer

To solve the given problem step by step, we will break down the solution into manageable parts.

### Step 1: Define the Determinant
We start with the determinant defined as:
\[ \Delta_n = \begin{vmatrix} a^2 + n & ab & ac \\ ab & b^2 + n & bc \\ ac & bc & c^2 + n \end{vmatrix} \]

### Step 2: Factor out common terms
We can factor out \(a\), \(b\), and \(c\) from the first, second, and third rows respectively:
\[ \Delta_n = abc \begin{vmatrix} 1 & \frac{ab}{a} & \frac{ac}{a} \\ \frac{ab}{b} & 1 & \frac{bc}{b} \\ \frac{ac}{c} & \frac{bc}{c} & 1 \end{vmatrix} \]
This simplifies to:
\[ \Delta_n = abc \begin{vmatrix} 1 & b & c \\ a & 1 & c \\ a & b & 1 \end{vmatrix} \]

### Step 3: Calculate the determinant
We can calculate the determinant using the formula for a 3x3 matrix:
\[ \Delta_n = abc \left( 1(1 \cdot 1 - b \cdot c) - b(a \cdot 1 - c \cdot a) + c(a \cdot b - 1 \cdot a) \right) \]
This simplifies to:
\[ \Delta_n = abc \left( 1 - bc - ab + ac + ac - ab \right) \]
After simplification, we find:
\[ \Delta_n = abc \left( a^2 + b^2 + c^2 + n \right) \]

### Step 4: Use the roots of the cubic equation
Given the cubic equation:
\[ x^3 - \lambda x^2 + 11x - 6 = 0 \]
with roots \(a\), \(b\), and \(c\) in arithmetic progression (AP). Since \(a\), \(b\), and \(c\) are in AP, we can express them as:
\[ b = a + d, \quad c = a + 2d \]
From Vieta's formulas, we know:
\[ a + b + c = \lambda, \quad ab + ac + bc = 11, \quad abc = 6 \]

### Step 5: Find \( \lambda \)
Substituting \(b\) and \(c\) into \(a + b + c = \lambda\):
\[ a + (a + d) + (a + 2d) = 3a + 3d = \lambda \implies \lambda = 3(a + d) \]

### Step 6: Substitute into the determinant
Now substituting \(a\), \(b\), and \(c\) into \(\Delta_n\):
\[ \Delta_n = n^3 + n^2(a^2 + b^2 + c^2) \]
Using the expression for \(a^2 + b^2 + c^2\) in terms of \(d\):
\[ a^2 + b^2 + c^2 = 3a^2 + 3d^2 = 3(a^2 + d^2) \]
Thus:
\[ \Delta_n = n^3 + 14n^2 \]

### Step 7: Calculate the summation
We need to compute:
\[ \sum_{r=1}^{30} \frac{27\Delta_r - \Delta_{3r}}{27r^2} \]
Calculating \(\Delta_r\) and \(\Delta_{3r}\):
\[ \Delta_r = r^3 + 14r^2, \quad \Delta_{3r} = (3r)^3 + 14(3r)^2 = 27r^3 + 126r^2 \]
Substituting these into the summation:
\[ 27\Delta_r - \Delta_{3r} = 27(r^3 + 14r^2) - (27r^3 + 126r^2) = 378r^2 \]
Thus:
\[ \sum_{r=1}^{30} \frac{378r^2}{27r^2} = \sum_{r=1}^{30} 14 = 14 \times 30 = 420 \]

### Final Answer
The value of the summation is:
\[ \boxed{420} \]