To determine the tension for the elastic spring's length of $$(3a - 2b)$$, we can use Hooke's law which states that the force exerted by a spring is proportional to the extension or compression of the spring from its natural length. Mathematically, Hooke's law is given by:

$$ F = k \cdot \Delta x $$

where:

• $$F$$ is the force exerted by the spring.
• $$k$$ is the spring constant (a measure of the stiffness of the spring).
• $$\Delta x$$ is the displacement from the natural length.

Let's denote the natural (unstretched) length of the spring as $$l_0$$. Given that the spring's length is $$a$$ under a tension of $$3 \mathrm{~N}$$, and its length is $$b$$ under a tension of $$2 \mathrm{~N}$$, we can write the equations as:

$$ 3 \mathrm{~N} = k \cdot (a - l_0) $$

$$ 2 \mathrm{~N} = k \cdot (b - l_0) $$

Next, we need to find the length displacement $$(3a - 2b)$$ in terms of the natural length $$l_0$$. We express the displacements as:

$$ x = (3a - 2b) - l_0 $$

To solve for this, let's express $$a - l_0$$ and $$b - l_0$$ from the given conditions:

$$ a - l_0 = \frac{3 \mathrm{~N}}{k} $$

$$ b - l_0 = \frac{2 \mathrm{~N}}{k} $$

Substitute these into the displacement equation:

$$ x = (3a - 2b) - l_0 = 3 \left( \frac{3 \mathrm{~N}}{k} + l_0 \right) - 2 \left( \frac{2 \mathrm{~N}}{k} + l_0 \right) - l_0 $$

On simplifying, we get:

$$ x = 3 \frac{3 \mathrm{~N}}{k} + 3 l_0 - 2 \frac{2 \mathrm{~N}}{k} - 2 l_0 - l_0 $$

$$ x = \frac{9 \mathrm{~N}}{k} + 3 l_0 - \frac{4 \mathrm{~N}}{k} - 3 l_0 $$

$$ x = \frac{5 \mathrm{~N}}{k} $$

Finally, by Hooke's Law, the force corresponding to this displacement is:

$$ F = k \cdot x = k \cdot \frac{5 \mathrm{~N}}{k} = 5 \mathrm{~N} $$

Therefore, the tension required for the elastic spring's length $$(3a - 2b)$$ is 5 N.