Let det `A=|{:(l,m,n),(p,q,r),(1,1,1):}|` and 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 `("det." A)^2` equals :

Options

A. 36
B. 100
C. 144
D. 160

Answer

To solve the problem, we need to calculate the value of \((\text{det} A)^2\) given the determinant \(A\) and the conditions provided.

Given:
\[ \text{det} A = \begin{vmatrix} l & m & n \\ p & q & r \\ 1 & 1 & 1 \end{vmatrix} \]
We also have the following equations:
1. \((l - m)^2 + (p - q)^2 = 9\)
2. \((m - n)^2 + (q - r)^2 = 16\)
3. \((n - l)^2 + (r - p)^2 = 25\)

### Step 1: Interpret the conditions as distances between points
The equations represent the squared distances between points in a coordinate system:
- Let \(A(l, m)\), \(B(p, q)\), and \(C(n, r)\) be points in a plane.
- The first equation represents the distance \(AB\), the second \(BC\), and the third \(CA\).

### Step 2: Identify the lengths of the sides of the triangle
From the equations:
- \(AB^2 = 9 \implies AB = 3\)
- \(BC^2 = 16 \implies BC = 4\)
- \(CA^2 = 25 \implies CA = 5\)

### Step 3: Use Heron's formula to find the area of the triangle
To find the area of triangle \(ABC\), we can use Heron's formula. First, we calculate the semi-perimeter \(s\):
\[ s = \frac{AB + BC + CA}{2} = \frac{3 + 4 + 5}{2} = 6 \]

Now, apply Heron's formula:
\[ \text{Area} = \sqrt{s(s - AB)(s - BC)(s - CA)} \]
Substituting the values:
\[ \text{Area} = \sqrt{6(6 - 3)(6 - 4)(6 - 5)} = \sqrt{6 \cdot 3 \cdot 2 \cdot 1} = \sqrt{36} = 6 \]

### Step 4: Relate the area to the determinant
The area of the triangle can also be expressed in terms of the determinant:
\[ \text{Area} = \frac{1}{2} \cdot \text{det} A \]
Thus, we have:
\[ 6 = \frac{1}{2} \cdot \text{det} A \implies \text{det} A = 12 \]

### Step 5: Calculate \((\text{det} A)^2\)
Finally, we find:
\[ (\text{det} A)^2 = 12^2 = 144 \]

### Conclusion
The value of \((\text{det} A)^2\) is:
\[ \boxed{144} \]