If x, y and z are real numbers such that ` ( x - 3) ^(2) + ( y - 4) ^(2) + ( z - 5)^(2) = 0 ` then , (x + y + z) is equal to

Options

A. `-12`
B. 0
C. 8
D. 12

Answer

To solve the equation \( (x - 3)^2 + (y - 4)^2 + (z - 5)^2 = 0 \), we can follow these steps:

### Step 1: Understand the equation
The equation consists of three squared terms. Since the square of any real number is always non-negative, the only way for their sum to equal zero is if each individual term is also zero.

### Step 2: Set each term to zero
We can set each squared term to zero:
1. \( (x - 3)^2 = 0 \)
2. \( (y - 4)^2 = 0 \)
3. \( (z - 5)^2 = 0 \)

### Step 3: Solve each equation
Now, we will solve each of these equations:
1. From \( (x - 3)^2 = 0 \):
\[ x - 3 = 0 \implies x = 3 \]
2. From \( (y - 4)^2 = 0 \):
\[ y - 4 = 0 \implies y = 4 \]
3. From \( (z - 5)^2 = 0 \):
\[ z - 5 = 0 \implies z = 5 \]

### Step 4: Calculate \( x + y + z \)
Now that we have the values of \( x \), \( y \), and \( z \):
\[ x + y + z = 3 + 4 + 5 \]

### Step 5: Perform the addition
Calculating the sum:
\[ 3 + 4 = 7 \]
\[ 7 + 5 = 12 \]

### Conclusion
Thus, \( x + y + z = 12 \).

### Final Answer
The value of \( x + y + z \) is \( 12 \).

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