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 find the least possible value of the determinant

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

where \( a, b, c, \lambda \in \mathbb{N} \), we can follow these steps:

### Step 1: Rewrite the Determinant
We start with the determinant:

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

### Step 2: Apply Row Operations
We can perform row operations to simplify the determinant. Let's add the first row to the second and third rows:

\[ R_2 \rightarrow R_2 + R_1 \quad \text{and} \quad R_3 \rightarrow R_3 + R_1 \]

This gives us:

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

### Step 3: Factor Out Common Terms
Now, we can factor out \( a^2 + b^2 + c^2 + \lambda \) from the first row:

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

### Step 4: Further Simplification
Next, we can simplify the remaining determinant by performing column operations. Subtract the first column from the second and third columns:

\[ C_2 \rightarrow C_2 - C_1 \quad \text{and} \quad C_3 \rightarrow C_3 - C_1 \]

This results in:

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

### Step 5: Calculate the Determinant
Now we can compute the determinant. The minimum value occurs when \( a = b = c = \lambda = 1 \):

\[ D_{\text{min}} = (1^2 + 1^2 + 1^2 + 1) \cdot \text{determinant of the simplified matrix} \]

Calculating this gives:

\[ D_{\text{min}} = 4 \cdot 1 = 4 \]

### Conclusion
Thus, the least possible value of the determinant is:

\[ \boxed{4} \]