Find the position vector of M.

M is the midpoint of XY. The position vector of M, $\vec{OM}$, is given by:

$$\vec{OM} = \frac{1}{2}(\vec{OX} + \vec{OY})$$

Substitute the given vectors $\vec{OX} = -4\vec{i} + 6\vec{j}$ and $\vec{OY} = 12\vec{i} - 6\vec{j}$:

$$\vec{OM} = \frac{1}{2}((-4\vec{i} + 6\vec{j}) + (12\vec{i} - 6\vec{j}))$$ $$\vec{OM} = \frac{1}{2}((-4+12)\vec{i} + (6-6)\vec{j})$$ $$\vec{OM} = \frac{1}{2}(8\vec{i} + 0\vec{j})$$ $$\vec{OM} = 4\vec{i} + 0\vec{j}$$

Given that ON:NY = 1:2, N divides the line segment OY in the ratio 1:2. Therefore, the position vector of N, $\vec{ON}$, is:

$$\vec{ON} = \frac{1}{1+2}\vec{OY} = \frac{1}{3}\vec{OY}$$

Substitute $\vec{OY} = 12\vec{i} - 6\vec{j}$:

$$\vec{ON} = \frac{1}{3}(12\vec{i} - 6\vec{j})$$ $$\vec{ON} = (\frac{1}{3} \times 12)\vec{i} - (\frac{1}{3} \times 6)\vec{j}$$ $$\vec{ON} = 4\vec{i} - 2\vec{j}$$

The vector $\vec{NM}$ is found by subtracting the position vector of N from the position vector of M:

$$\vec{NM} = \vec{OM} - \vec{ON}$$

Substitute the calculated $\vec{OM}$ and $\vec{ON}$:

$$\vec{NM} = (4\vec{i} + 0\vec{j}) - (4\vec{i} - 2\vec{j})$$ $$\vec{NM} = (4-4)\vec{i} + (0 - (-2))\vec{j}$$ $$\vec{NM} = 0\vec{i} + 2\vec{j}$$

The magnitude of a vector $a\vec{i} + b\vec{j}$ is given by $\sqrt{a^2 + b^2}$.

$$|\vec{NM}| = \sqrt{0^2 + 2^2}$$ $$|\vec{NM}| = \sqrt{0 + 4}$$ $$|\vec{NM}| = \sqrt{4}$$ $$|\vec{NM}| = 2$$

Correct to 1 decimal place, the magnitude is 2.0.

The magnitude of NM is $\text{2.0 units}$.