To solve the given problem, we will break it down into steps based on the conditions provided.
### Step 1: Analyze the first equation
Given the equation:
\[
(d + a - b)^2 + (d + b - c)^2 = 0
\]
Since the sum of two squares is zero, both squares must individually be zero. Therefore, we can set up the following equations:
1. \(d + a - b = 0\)
2. \(d + b - c = 0\)
### Step 2: Solve for \(d\)
From the first equation:
\[
d = b - a
\]
From the second equation:
\[
d = c - b
\]
Since both expressions equal \(d\), we can set them equal to each other:
\[
b - a = c - b
\]
### Step 3: Rearrange the equation
Rearranging the equation gives:
\[
2b = a + c
\]
This implies:
\[
b = \frac{a + c}{2}
\]
This shows that \(a\), \(b\), and \(c\) are in Arithmetic Progression (AP).
### Step 4: Analyze the second equation
The second equation given is:
\[
a(b - c)x^2 + b(c - a)x + c(a - b) = 0
\]
For the roots to be real and equal, the discriminant must be zero:
\[
B^2 - 4AC = 0
\]
Here, \(A = a(b - c)\), \(B = b(c - a)\), and \(C = c(a - b)\).
### Step 5: Calculate the discriminant
Calculating the discriminant:
\[
[b(c - a)]^2 - 4[a(b - c)][c(a - b)] = 0
\]
Expanding this gives:
\[
b^2(c - a)^2 - 4ac(b - c)(a - b) = 0
\]
### Step 6: Substitute \(b\) in terms of \(a\) and \(c\)
Substituting \(b = \frac{a + c}{2}\) into the discriminant:
1. Calculate \(b - c\) and \(c - a\):
\[
b - c = \frac{a + c}{2} - c = \frac{a - c}{2}
\]
\[
c - a = c - a
\]
### Step 7: Simplify the discriminant condition
Substituting these values into the discriminant condition and simplifying will lead to relationships among \(a\), \(b\), and \(c\).
### Step 8: Check for GP condition
From the discriminant condition, we can derive that:
\[
b^2 = ac
\]
This shows that \(a\), \(b\), and \(c\) are in Geometric Progression (GP).
### Step 9: Check for HP condition
Since we have established that \(b = \frac{a + c}{2}\), we can also show that:
\[
\frac{1}{b} = \frac{2}{a + c}
\]
This implies that \(a\), \(b\), and \(c\) are in Harmonic Progression (HP).
### Conclusion
Thus, we conclude that:
1. \(a\), \(b\), and \(c\) are in AP.
2. \(a\), \(b\), and \(c\) are in GP.
3. \(a\), \(b\), and \(c\) are in HP.
