Find the resultant force F_net by adding the three given forces.

The forces are given as:
$\vec{F}_1 = (2m + n)$
$\vec{F}_2 = (-3m + 4n)$
$\vec{F}_3 = (4m - 6n)$

We sum the components:

$$\vec{F}_{\text{net}} = \vec{F}_1 + \vec{F}_2 + \vec{F}_3$$ $$\vec{F}_{\text{net}} = (2m + n) + (-3m + 4n) + (4m - 6n)$$

Combine the $m$ components and the $n$ components:

$$\vec{F}_{\text{net}} = (2 - 3 + 4)m + (1 + 4 - 6)n$$ $$\vec{F}_{\text{net}} = 3m - n$$

Assuming $m$ and $n$ are orthogonal unit vectors, the magnitude is found using the Pythagorean theorem:

$$|\vec{F}_{\text{net}}| = \sqrt{(3)^2 + (-1)^2}$$ $$|\vec{F}_{\text{net}}| = \sqrt{9 + 1}$$ $$|\vec{F}_{\text{net}}| = \sqrt{10} \text{ N}$$

Newton's second law states that $\vec{F}_{\text{net}} = M\vec{a}$, where $M$ is the mass and $\vec{a}$ is the acceleration.
We are given the mass $M = 12 \text{ kg}$.
The magnitude of acceleration is given by:

$$|\vec{a}| = \frac{|\vec{F}_{\text{net}}|}{M}$$

Substitute the values:

$$|\vec{a}| = \frac{\sqrt{10} \text{ N}}{12 \text{ kg}}$$ $$|\vec{a}| = \frac{\sqrt{10}}{12} \text{ m/s}^2$$

The magnitude of the acceleration of the body is $\frac{\sqrt{10}}{12} \text{ m/s}^2$.