We can solve this using Hooke's Law, which states:

$$F = kx,$$

where:

• $$F$$ is the force (tension),

• $$k$$ is the spring constant, and

• $$x$$ is the extension of the spring.

Follow these steps:

For the first scenario (extension $$x_1$$ under 5 N):

$$5 = kx_1 \quad \Rightarrow \quad k = \frac{5}{x_1}.$$

For the second scenario (extension $$x_2$$ under 7 N):

$$7 = kx_2 \quad \Rightarrow \quad k = \frac{7}{x_2}.$$

Equate the two expressions for $$k$$:

$$\frac{5}{x_1} = \frac{7}{x_2} \quad \Rightarrow \quad \frac{x_1}{x_2} = \frac{5}{7} \quad \Rightarrow \quad x_2 = \frac{7}{5}x_1.$$

To find the tension for the extension of $$\left(5x_1 - 2x_2\right)$$, use Hooke's Law again:

$$\text{Tension, } F = k \left(5x_1 - 2x_2\right).$$

Substitute $$k = \frac{5}{x_1}$$:

$$F = \frac{5}{x_1} \left(5x_1 - 2x_2\right).$$

Substitute the expression for $$x_2$$:

$$F = \frac{5}{x_1} \left(5x_1 - 2\left(\frac{7}{5}x_1\right)\right) = \frac{5}{x_1} \left(5x_1 - \frac{14}{5}x_1\right).$$

Simplify the expression inside the parentheses:

$$5x_1 - \frac{14}{5}x_1 = \left(\frac{25}{5} - \frac{14}{5}\right)x_1 = \frac{11}{5}x_1.$$

Now substitute back:

$$F = \frac{5}{x_1} \cdot \frac{11}{5}x_1 = 11 \text{ N}.$$

Thus, the tension in the spring for the extension $$\left(5x_1-2x_2\right)$$ is $$11 \text{ N}.$$

The correct option is Option C.