Step 1: Understanding the Concept:
The given determinant represents twice the area of a triangle formed by the vertices \( (l, p), (m, q), \) and \( (n, r) \). The provided squared differences are the squared distances between these vertices.
Step 2: Key Formula or Approach:
1. Let the vertices be \( A(l, p), B(m, q), \) and \( C(n, r) \).
2. Side lengths squared: \( AB^2 = 9, BC^2 = 16, CA^2 = 25 \).
3. Side lengths: \( AB = 3, BC = 4, CA = 5 \).
4. The area of a triangle with vertices \( (x_1, y_1), (x_2, y_2), (x_3, y_3) \) is \( \frac{1}{2} | \text{det } A | \).
Step 3: Detailed Explanation:
The side lengths \( 3, 4, 5 \) form a Pythagorean triplet, so the triangle is a right-angled triangle.
Area of triangle \( ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 3 \times 4 = 6 \).
We know that for vertices \( (l, p), (m, q), (n, r) \):
\[ \text{Area} = \frac{1}{2} \left| \begin{vmatrix} l & p & 1
m & q & 1
n & r & 1 \end{vmatrix} \right| \]
The given determinant \( \text{det } A \) is:
\[ \begin{vmatrix} l & m & n
p & q & r
1 & 1 & 1 \end{vmatrix} \]
By taking the transpose, we get the vertical form. Transpose doesn't change the determinant value.
\[ \begin{vmatrix} l & p & 1
m & q & 1
n & r & 1 \end{vmatrix} = \text{det } A \]
So, \( \text{Area} = \frac{1}{2} |\text{det } A| \).
Substitute Area = 6:
\[ 6 = \frac{1}{2} |\text{det } A| \implies |\text{det } A| = 12 \]
The question asks for \( (\text{det } A)^2 \):
\[ (\text{det } A)^2 = 12^2 = 144 \]
Step 4: Final Answer:
The value of \( (\text{det } A)^2 \) is 144.