If the system of equation `14x-3y+z=12, x-2y=0 and x+2z=0` has a solution `(x_(0), y_(0), z_(0))`, then the value of `x_(0)^(2)+y_(2)^(2)+z_(2)^(2)` is equal to

Options

A. `(3)/(2)`
B. `(3)/(4)`
C. `(9)/(2)`
D. `(9)/(4)`

Answer

To solve the system of equations:

1. **Equations Given**:
\[ 14x - 3y + z = 12 \quad \text{(Equation 1)} \]
\[ x - 2y = 0 \quad \text{(Equation 2)} \]
\[ x + 2z = 0 \quad \text{(Equation 3)} \]

2. **From Equation 2**:
Rearranging Equation 2 gives:
\[ x = 2y \quad \text{(Equation 4)} \]

3. **From Equation 3**:
Rearranging Equation 3 gives:
\[ z = -\frac{x}{2} \quad \text{(Equation 5)} \]

4. **Substituting Equation 4 into Equation 5**:
Substitute \(x = 2y\) into Equation 5:
\[ z = -\frac{2y}{2} = -y \quad \text{(Equation 6)} \]

5. **Substituting Equations 4 and 6 into Equation 1**:
Now substitute \(x = 2y\) and \(z = -y\) into Equation 1:
\[ 14(2y) - 3y + (-y) = 12 \]
Simplifying this gives:
\[ 28y - 3y - y = 12 \]
\[ 24y = 12 \]
\[ y = \frac{12}{24} = \frac{1}{2} \]

6. **Finding x and z**:
Substitute \(y = \frac{1}{2}\) back into Equation 4 to find \(x\):
\[ x = 2\left(\frac{1}{2}\right) = 1 \]
Substitute \(y = \frac{1}{2}\) into Equation 6 to find \(z\):
\[ z = -\left(\frac{1}{2}\right) = -\frac{1}{2} \]

7. **Values of \(x_0\), \(y_0\), and \(z_0\)**:
Thus, we have:
\[ (x_0, y_0, z_0) = \left(1, \frac{1}{2}, -\frac{1}{2}\right) \]

8. **Calculating \(x_0^2 + y_0^2 + z_0^2\)**:
Now calculate:
\[ x_0^2 + y_0^2 + z_0^2 = 1^2 + \left(\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right)^2 \]
\[ = 1 + \frac{1}{4} + \frac{1}{4} \]
\[ = 1 + \frac{2}{4} = 1 + \frac{1}{2} = \frac{3}{2} \]

Thus, the value of \(x_0^2 + y_0^2 + z_0^2\) is \(\frac{3}{2}\).