Consider the following statements in respect of the determinant Δ =...

Given determinant is $\Delta = \begin{vmatrix} k(k+2) \text{ and } 2k+1 \text{ and } 1 \\ 2k+1 \text{ and } k+2 \text{ and } 1 \\ 3 \text{ and } 3 \text{ and } 1 \end{vmatrix}$

Applying row operations $R_1 \rightarrow R_1 - R_2$ and $R_2 \rightarrow R_2 - R_3$:

$\Delta = \begin{vmatrix} k^2-1 \text{ and } k-1 \text{ and } 0 \\ 2k-2 \text{ and } k-1 \text{ and } 0 \\ 3 \text{ and } 3 \text{ and } 1 \end{vmatrix}$

Expanding along the third column:

$\Delta = 1 \cdot \begin{vmatrix} k^2-1 \text{ and } k-1 \\ 2k-2 \text{ and } k-1 \end{vmatrix}$

$\Delta = (k^2-1)(k-1) - (2k-2)(k-1)$

$\Delta = (k-1) [ k^2 - 1 - 2k + 2 ]$

$\Delta = (k-1) (k^2 - 2k + 1)$

$\Delta = (k-1)^3$

Evaluating the given statements:

I. For $k = \dfrac{1}{2} > 0$, $\Delta = \left(\dfrac{1}{2} - 1\right)^3 = -\dfrac{1}{8} < 0$. Thus, statement I is incorrect.

II. For $k < 0$, $k - 1 < -1$, which implies $(k-1)^3 < 0$. Thus, statement II is correct.

III. For $k = 0$, $\Delta = (0-1)^3 = -1 \neq 0$. Thus, statement III is incorrect.

Only one statement is correct.

Answer: One