If `(x+ y-z -1)^(2) + (z+ x-y - 2)^(2) + (z+y-x-4)^(2)=0` find `x+ y+z`=?

Answer

To solve the equation

\[ (x + y - z - 1)^2 + (z + x - y - 2)^2 + (z + y - x - 4)^2 = 0, \]

we start by recognizing that each term in the equation is a square. Since the sum of squares is equal to zero, each individual square must also be equal to zero. This leads us to the following three equations:

1. \( x + y - z - 1 = 0 \)
2. \( z + x - y - 2 = 0 \)
3. \( z + y - x - 4 = 0 \)

Now, we will solve these equations step by step.

### Step 1: Solve the first equation
From the first equation, we can express \( z \) in terms of \( x \) and \( y \):

\[ x + y - z - 1 = 0 \implies z = x + y - 1. \]

### Step 2: Substitute \( z \) into the second equation
Now, substitute \( z \) into the second equation:

\[ z + x - y - 2 = 0 \implies (x + y - 1) + x - y - 2 = 0. \]

Simplifying this gives:

\[ x + y - 1 + x - y - 2 = 0 \implies 2x - 3 = 0 \implies 2x = 3 \implies x = \frac{3}{2}. \]

### Step 3: Substitute \( x \) back to find \( y \) and \( z \)
Now that we have \( x \), we can substitute \( x \) back into the expression for \( z \):

\[ z = x + y - 1 \implies z = \frac{3}{2} + y - 1 \implies z = y + \frac{1}{2}. \]

Next, we substitute \( x \) into the third equation:

\[ z + y - x - 4 = 0 \implies (y + \frac{1}{2}) + y - \frac{3}{2} - 4 = 0. \]

Simplifying this gives:

\[ y + \frac{1}{2} + y - \frac{3}{2} - 4 = 0 \implies 2y - 4 = 0 \implies 2y = 4 \implies y = 2. \]

### Step 4: Find \( z \)
Now substitute \( y \) back into the equation for \( z \):

\[ z = y + \frac{1}{2} \implies z = 2 + \frac{1}{2} = \frac{5}{2}. \]

### Step 5: Calculate \( x + y + z \)
Now we have:

- \( x = \frac{3}{2} \)
- \( y = 2 \)
- \( z = \frac{5}{2} \)

Now we can find \( x + y + z \):

\[ x + y + z = \frac{3}{2} + 2 + \frac{5}{2} = \frac{3}{2} + \frac{4}{2} + \frac{5}{2} = \frac{12}{2} = 6. \]

Thus, the final answer is:

\[ x + y + z = 6. \]