If a,b,c are real numbers such that a+b+c=0 and `a^(2)+b^(2)+c^(2)=1,` then
`(3a+5b-8c)^(2)+(-8a+3b+5c)^(2)+(5a-8b+3c)^(2)` is equal to

Options

A. 49
B. 98
C. 147
D. 294

Explanation

To solve the problem, we need to evaluate the expression \((3a + 5b - 8c)^2 + (-8a + 3b + 5c)^2 + (5a - 8b + 3c)^2\) given the conditions \(a + b + c = 0\) and \(a^2 + b^2 + c^2 = 1\).

### Step 1: Expand Each Square
We will expand each term in the expression separately.

1. **First term:**
\[ (3a + 5b - 8c)^2 = 9a^2 + 25b^2 + 64c^2 + 30ab - 48ac - 80bc \]

2. **Second term:**
\[ (-8a + 3b + 5c)^2 = 64a^2 + 9b^2 + 25c^2 - 48ab + 30bc - 8ac \]

3. **Third term:**
\[ (5a - 8b + 3c)^2 = 25a^2 + 64b^2 + 9c^2 - 8ab - 48bc + 30ac \]

### Step 2: Combine the Expanded Terms
Now we will add all three expanded terms together:

\[ (3a + 5b - 8c)^2 + (-8a + 3b + 5c)^2 + (5a - 8b + 3c)^2 \]

Combining like terms:
- Coefficients of \(a^2\):
\[ 9a^2 + 64a^2 + 25a^2 = 98a^2 \]

- Coefficients of \(b^2\):
\[ 25b^2 + 9b^2 + 64b^2 = 98b^2 \]

- Coefficients of \(c^2\):
\[ 64c^2 + 25c^2 + 9c^2 = 98c^2 \]

- Coefficients of \(ab\):
\[ 30ab - 48ab - 8ab = -26ab \]

- Coefficients of \(ac\):
\[ -48ac - 8ac + 30ac = -26ac \]

- Coefficients of \(bc\):
\[ -80bc + 30bc - 48bc = -98bc \]

### Step 3: Write the Combined Expression
Putting it all together, we have:
\[ 98(a^2 + b^2 + c^2) - 26(ab + ac + bc) \]

### Step 4: Use Given Conditions
We know from the problem that:
- \(a + b + c = 0\)
- \(a^2 + b^2 + c^2 = 1\)

Using the identity:
\[ a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + ac + bc) \]
Substituting \(a + b + c = 0\):
\[ 1 = 0 - 2(ab + ac + bc) \implies ab + ac + bc = -\frac{1}{2} \]

### Step 5: Substitute Values
Now substitute \(a^2 + b^2 + c^2 = 1\) and \(ab + ac + bc = -\frac{1}{2}\) into the combined expression:
\[ 98(1) - 26\left(-\frac{1}{2}\right) = 98 + 13 = 111 \]

### Final Answer
Thus, the value of the expression is:
\[ \boxed{111} \]