Let $\det A = \begin{vmatrix} l & m & n \\ p & q & r \\ 1 & 1 & 1 \end{vmatrix}$. If $(l - m)^2 + (p - q)^2 = 9, (m - n)^2 + (q - r)^2 = 16, (n - l)^2 + (r - p)^2 = 25$, then the value of $(\det A)^2$ is

Let the points in the Cartesian plane be $P_1(l, p)$, $P_2(m, q)$, and $P_3(n, r)$.

The given equations represent the squares of the distances between these points:

$P_1 P_2^2 = (l - m)^2 + (p - q)^2 = 9 \Rightarrow P_1 P_2 = 3$

$P_2 P_3^2 = (m - n)^2 + (q - r)^2 = 16 \Rightarrow P_2 P_3 = 4$

$P_3 P_1^2 = (n - l)^2 + (r - p)^2 = 25 \Rightarrow P_3 P_1 = 5$

Since $P_1 P_2^2 + P_2 P_3^2 = 9 + 16 = 25 = P_3 P_1^2$, the triangle formed by these three points is a right-angled triangle.

The area of $\Delta P_1 P_2 P_3$ is given by:

Area $= \dfrac{1}{2} \times P_1 P_2 \times P_2 P_3 = \dfrac{1}{2} \times 3 \times 4 = 6$

The area of a triangle with vertices $(l, p)$, $(m, q)$, and $(n, r)$ can also be expressed using a determinant:

Area $= \dfrac{1}{2} \left| \begin{vmatrix} l & m & n \\ p & q & r \\ 1 & 1 & 1 \end{vmatrix} \right| = \dfrac{1}{2} |\det A|$

Equating the two expressions for the area:

$\dfrac{1}{2} |\det A| = 6 \Rightarrow |\det A| = 12$

Squaring both sides, we obtain:

$(\det A)^2 = 144$

Answer: $144$