If `(3a+1)^(2) +(b-1)^(2) +(2c-3)^(2)=0`, then value of `(3a+b+2c)` is

Options

A. 3
B. `-1`
C. 2
D. 5

Answer

To solve the equation \( (3a + 1)^2 + (b - 1)^2 + (2c - 3)^2 = 0 \), we need to analyze the components of the equation step by step.

### 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 zero.

### Step 2: Set each term to zero
We can set each squared term to zero:

1. \( (3a + 1)^2 = 0 \)
2. \( (b - 1)^2 = 0 \)
3. \( (2c - 3)^2 = 0 \)

### Step 3: Solve for \(a\), \(b\), and \(c\)

1. **For \( (3a + 1)^2 = 0 \)**:
\[ 3a + 1 = 0 \implies 3a = -1 \implies a = -\frac{1}{3} \]

2. **For \( (b - 1)^2 = 0 \)**:
\[ b - 1 = 0 \implies b = 1 \]

3. **For \( (2c - 3)^2 = 0 \)**:
\[ 2c - 3 = 0 \implies 2c = 3 \implies c = \frac{3}{2} \]

### Step 4: Calculate \(3a + b + 2c\)
Now that we have the values of \(a\), \(b\), and \(c\), we can substitute them into the expression \(3a + b + 2c\):

\[ 3a + b + 2c = 3\left(-\frac{1}{3}\right) + 1 + 2\left(\frac{3}{2}\right) \]

Calculating each term:

- \(3 \times -\frac{1}{3} = -1\)
- \(b = 1\)
- \(2 \times \frac{3}{2} = 3\)

Now, combine these results:

\[ 3a + b + 2c = -1 + 1 + 3 \]

### Step 5: Simplify the expression
\[ -1 + 1 + 3 = 0 + 3 = 3 \]

### Final Answer
Thus, the value of \(3a + b + 2c\) is \(3\).

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