If `a,b, c, lambda in N`, then the least possible value of `|(a^(2)+lambda, ab,ac),(ba,b^(2)+lambda,bc),(ca, cb, c^(2)+lambda)|` is

Answer

To solve the problem, we need to find the least possible value of the determinant:

\[ D = \begin{vmatrix} a^2 + \lambda & ab & ac \\ ba & b^2 + \lambda & bc \\ ca & cb & c^2 + \lambda \end{vmatrix} \]

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

\[ D = \begin{vmatrix} a(a^2 + \lambda) & a(ab) & a(ac) \\ b(ba) & b(b^2 + \lambda) & b(bc) \\ c(ca) & c(cb) & c(c^2 + \lambda) \end{vmatrix} \]

This simplifies to:

\[ D = \begin{vmatrix} a^3 + a\lambda & a^2b & a^2c \\ b^2a & b^3 + b\lambda & b^2c \\ c^2a & c^2b & c^3 + c\lambda \end{vmatrix} \]

### Step 2: Factor Out Common Terms
Next, we can factor out \(a\), \(b\), and \(c\) from the first, second, and third columns respectively:

\[ D = abc \begin{vmatrix} a^2 + \lambda & ab & ac \\ ba & b^2 + \lambda & bc \\ ca & cb & c^2 + \lambda \end{vmatrix} \]

### Step 3: Row Operation
Now, we perform the row operation \(R_1 \to R_1 + R_2 + R_3\):

\[ D = abc \begin{vmatrix} a^2 + b^2 + c^2 + \lambda & ab & ac \\ ba & b^2 + \lambda & bc \\ ca & cb & c^2 + \lambda \end{vmatrix} \]

### Step 4: Factor Out Common Terms Again
We can factor out \(a^2 + b^2 + c^2 + \lambda\) from the first row:

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

### Step 5: Simplify the Determinant
Now we can simplify the determinant further by performing row operations to make as many entries zero as possible. We can subtract the first row from the second and third rows:

\[ D = abc (a^2 + b^2 + c^2 + \lambda) \begin{vmatrix} 1 & 1 & 1 \\ 0 & \frac{\lambda}{b} & 0 \\ 0 & 0 & \frac{\lambda}{c} \end{vmatrix} \]

### Step 6: Calculate the Determinant
The determinant simplifies to:

\[ D = abc (a^2 + b^2 + c^2 + \lambda) \cdot \frac{\lambda^2}{bc} \]

### Step 7: Find the Minimum Value
To find the minimum value, we set \(a = 1\), \(b = 1\), \(c = 1\), and \(\lambda = 1\):

\[ D = 1 \cdot 1 \cdot 1 \cdot (1^2 + 1^2 + 1^2 + 1) \cdot \frac{1^2}{1} = 1 \cdot 4 = 4 \]

Thus, the least possible value of the determinant is:

\[ \boxed{4} \]