To solve the problem, we need to find the value of \((AB^2 + BC^2 + CA^2) / (l^2 + m^2 + n^2)\) given the midpoints of the sides of triangle \(ABC\) as \((l, 0, 0)\), \((0, m, 0)\), and \((0, 0, n)\).
### Step 1: Establish the Coordinates of Points A, B, and C
Let the coordinates of points \(A\), \(B\), and \(C\) be:
- \(A = (x_1, y_1, z_1)\)
- \(B = (x_2, y_2, z_2)\)
- \(C = (x_3, y_3, z_3)\)
### Step 2: Use the Midpoint Formula
The midpoints of the sides are given as:
- Midpoint of \(AB = (l, 0, 0)\)
- Midpoint of \(BC = (0, m, 0)\)
- Midpoint of \(CA = (0, 0, n)\)
Using the midpoint formula:
1. For \(AB\):
\[
\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2}\right) = (l, 0, 0)
\]
This gives us:
\[
x_1 + x_2 = 2l, \quad y_1 + y_2 = 0, \quad z_1 + z_2 = 0
\]
2. For \(BC\):
\[
\left(\frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2}, \frac{z_2 + z_3}{2}\right) = (0, m, 0)
\]
This gives us:
\[
x_2 + x_3 = 0, \quad y_2 + y_3 = 2m, \quad z_2 + z_3 = 0
\]
3. For \(CA\):
\[
\left(\frac{x_3 + x_1}{2}, \frac{y_3 + y_1}{2}, \frac{z_3 + z_1}{2}\right) = (0, 0, n)
\]
This gives us:
\[
x_3 + x_1 = 0, \quad y_3 + y_1 = 0, \quad z_3 + z_1 = 2n
\]
### Step 3: Solve the System of Equations
From the equations derived:
1. From \(x_2 + x_3 = 0\), we have \(x_3 = -x_2\).
2. Substituting \(x_3\) into \(x_1 + x_2 = 2l\) gives:
\[
x_1 - x_2 = 2l \implies x_1 = 2l + x_2
\]
3. Substituting \(x_1\) into \(x_3 + x_1 = 0\) gives:
\[
-x_2 + (2l + x_2) = 0 \implies 2x_2 = -2l \implies x_2 = -l \implies x_1 = l, x_3 = l
\]
Following similar steps for \(y\) and \(z\):
- For \(y\):
\[
y_1 = -m, \quad y_2 = m, \quad y_3 = m
\]
- For \(z\):
\[
z_1 = n, \quad z_2 = -n, \quad z_3 = n
\]
### Step 4: Determine the Coordinates of A, B, and C
Thus, the coordinates are:
- \(A = (l, -m, n)\)
- \(B = (-l, m, -n)\)
- \(C = (-l, m, n)\)
### Step 5: Calculate the Distances
Now we calculate \(AB^2\), \(BC^2\), and \(CA^2\):
1. \(AB^2\):
\[
AB^2 = (l - (-l))^2 + (-m - m)^2 + (n - (-n))^2 = (2l)^2 + (-2m)^2 + (2n)^2 = 4l^2 + 4m^2 + 4n^2
\]
2. \(BC^2\):
\[
BC^2 = (-l - (-l))^2 + (m - m)^2 + (-n - n)^2 = 0 + 0 + (-2n)^2 = 4n^2
\]
3. \(CA^2\):
\[
CA^2 = (-l - l)^2 + (m - (-m))^2 + (n - n)^2 = (-2l)^2 + (2m)^2 + 0 = 4l^2 + 4m^2
\]
### Step 6: Combine the Results
Now we combine these results:
\[
AB^2 + BC^2 + CA^2 = (4l^2 + 4m^2 + 4n^2) + (4n^2) + (4l^2 + 4m^2) = 8(l^2 + m^2 + n^2)
\]
### Step 7: Final Calculation
Now we compute:
\[
\frac{AB^2 + BC^2 + CA^2}{l^2 + m^2 + n^2} = \frac{8(l^2 + m^2 + n^2)}{l^2 + m^2 + n^2} = 8
\]
### Final Answer
Thus, the value is \(8\).
