The sum of squares of three positive integers is 323. If the sum of squares of two numbers is twice the third, their product is

Options

A. 255
B. 260
C. 265
D. 270

Answer

To solve the problem, we need to find three positive integers \( a \), \( b \), and \( c \) such that:

1. The sum of their squares is 323:
\[ a^2 + b^2 + c^2 = 323 \]

2. The sum of the squares of two of these numbers is twice the square of the third:
\[ a^2 + b^2 = 2c^2 \]

### Step-by-Step Solution:

**Step 1: Set up the equations.**

From the given information, we have:
1. \( a^2 + b^2 + c^2 = 323 \)
2. \( a^2 + b^2 = 2c^2 \)

**Step 2: Substitute \( a^2 + b^2 \) in the first equation.**

From the second equation, we can express \( a^2 + b^2 \) in terms of \( c^2 \):
\[ a^2 + b^2 = 2c^2 \]

Now, substitute this into the first equation:
\[ 2c^2 + c^2 = 323 \]
This simplifies to:
\[ 3c^2 = 323 \]

**Step 3: Solve for \( c^2 \).**

Now, divide both sides by 3:
\[ c^2 = \frac{323}{3} \approx 107.67 \]
Since \( c^2 \) must be an integer, we check for integer values of \( c \).

**Step 4: Find integer values for \( c \).**

We can check the integer values of \( c \) that yield a valid \( c^2 \):
- If \( c = 17 \), then \( c^2 = 289 \).

**Step 5: Calculate \( a^2 + b^2 \).**

Substituting \( c = 17 \) back into the equation \( a^2 + b^2 = 2c^2 \):
\[ a^2 + b^2 = 2(289) = 578 \]

**Step 6: Substitute back into the first equation.**

Now, substituting \( c^2 = 289 \) into the first equation:
\[ a^2 + b^2 + 289 = 323 \]
This simplifies to:
\[ a^2 + b^2 = 323 - 289 = 34 \]

**Step 7: Find integers \( a \) and \( b \).**

Now we need to find two integers \( a \) and \( b \) such that:
\[ a^2 + b^2 = 34 \]
The pairs that satisfy this are:
- \( (5, 3) \) since \( 5^2 + 3^2 = 25 + 9 = 34 \).

**Step 8: Calculate the product \( a \times b \times c \).**

Now we can find the product:
\[ a \times b \times c = 5 \times 3 \times 17 = 255 \]

### Final Answer:
The product of the three integers \( a \), \( b \), and \( c \) is \( 255 \).