The number of solutions of the equation
`3x-y-z=0`
`-3x+2y+z=0`
`-3x+z=0`
such that x,y,z are non negative integers and `x^(2)+y^(2)+z^(2)le81` is

Options

A. 3
B. 7
C. 1
D. 2

Answer

To find the number of solutions of the equations:

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

such that \(x, y, z\) are non-negative integers and \(x^2 + y^2 + z^2 \leq 81\), we will follow these steps:

### Step 1: Solve the equations for \(y\) and \(z\) in terms of \(x\)

From the third equation, we can express \(z\) in terms of \(x\):

\[ z = 3x \]

### Step 2: Substitute \(z\) into the first and second equations

Now, substituting \(z = 3x\) into the first equation:

\[ 3x - y - 3x = 0 \implies y = 0 \]

Now substituting \(y = 0\) and \(z = 3x\) into the second equation:

\[ -3x + 2(0) + 3x = 0 \implies 0 = 0 \]

This means that the second equation is satisfied for any \(x\).

### Step 3: Determine the conditions for \(x\)

Now we have \(y = 0\) and \(z = 3x\). We need to ensure that \(x^2 + y^2 + z^2 \leq 81\):

\[ x^2 + 0^2 + (3x)^2 \leq 81 \]
\[ x^2 + 9x^2 \leq 81 \]
\[ 10x^2 \leq 81 \]
\[ x^2 \leq \frac{81}{10} = 8.1 \]
\[ x \leq \sqrt{8.1} \approx 2.84 \]

Since \(x\) must be a non-negative integer, the possible values for \(x\) are \(0, 1, 2\).

### Step 4: Calculate corresponding values of \(y\) and \(z\)

Now we can find the corresponding values of \(y\) and \(z\) for each integer value of \(x\):

- For \(x = 0\):
- \(y = 0\)
- \(z = 3(0) = 0\)
- Solution: \((0, 0, 0)\)

- For \(x = 1\):
- \(y = 0\)
- \(z = 3(1) = 3\)
- Solution: \((1, 0, 3)\)

- For \(x = 2\):
- \(y = 0\)
- \(z = 3(2) = 6\)
- Solution: \((2, 0, 6)\)

### Step 5: Check if there are any more solutions

For \(x = 3\):
- \(y = 0\)
- \(z = 3(3) = 9\)
- Checking the condition:
\[ 3^2 + 0^2 + 9^2 = 9 + 0 + 81 = 90 > 81 \]
This does not satisfy the condition.

For \(x \geq 3\), \(z\) will only increase, leading to \(x^2 + y^2 + z^2\) exceeding 81.

### Conclusion

The valid solutions are:
1. \((0, 0, 0)\)
2. \((1, 0, 3)\)
3. \((2, 0, 6)\)

Thus, the number of solutions is **3**.