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 find the value of \((\text{det } A)^2\) given the determinant of matrix \(A\) defined as:

\[ A = \begin{pmatrix} l & m & n \\ p & q & r \\ 1 & 1 & 1 \end{pmatrix} \]

And the conditions provided are:

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: Identify the lengths of the sides of the triangle

From the conditions, we can interpret them as the squares of the lengths of the sides of a triangle:

- Let \(AB\) be the distance between points \(A(l, p)\) and \(B(m, q)\):
\[ AB^2 = (l - m)^2 + (p - q)^2 = 9 \implies AB = 3 \]

- Let \(BC\) be the distance between points \(B(m, q)\) and \(C(n, r)\):
\[ BC^2 = (m - n)^2 + (q - r)^2 = 16 \implies BC = 4 \]

- Let \(CA\) be the distance between points \(C(n, r)\) and \(A(l, p)\):
\[ CA^2 = (n - l)^2 + (r - p)^2 = 25 \implies CA = 5 \]

### Step 2: Calculate the semi-perimeter of the triangle

The semi-perimeter \(s\) of triangle \(ABC\) is given by:
\[ s = \frac{AB + BC + CA}{2} = \frac{3 + 4 + 5}{2} = 6 \]

### Step 3: Calculate the area of triangle \(ABC\) using Heron's formula

Using Heron's formula, the area \(A\) of triangle \(ABC\) is:
\[ \text{Area} = \sqrt{s(s - AB)(s - BC)(s - CA)} \]
Substituting the values:
\[ \text{Area} = \sqrt{6 \times (6 - 3) \times (6 - 4) \times (6 - 5)} = \sqrt{6 \times 3 \times 2 \times 1} = \sqrt{36} = 6 \]

### Step 4: Relate the area to the determinant

The determinant of matrix \(A\) can be related to the area of triangle \(ABC\) by the formula:
\[ \text{det } A = 2 \times \text{Area} \]
Thus,
\[ \text{det } A = 2 \times 6 = 12 \]

### Step 5: Calculate \((\text{det } A)^2\)

Finally, we calculate \((\text{det } A)^2\):
\[ (\text{det } A)^2 = 12^2 = 144 \]

### Conclusion

The value of \((\text{det } A)^2\) is \(144\).