If the distance of the point $$(1, -2, 3)$$ from the plane $$x + 2y - 3z + 10 = 0$$ measured parallel to the line, $$\frac{x-1}{3} = \frac{2-y}{m} = \frac{z+3}{1}$$ is $$\sqrt{\frac{7}{2}}$$, then the value of $$|m|$$ is equal to ________.

The point is $$(1, -2, 3)$$ and the plane is $$x + 2y - 3z + 10 = 0$$. The distance is measured parallel to the line $$\frac{x-1}{3} = \frac{2-y}{m} = \frac{z+3}{1}$$, which has direction vector $$(3, -m, 1)$$.

A point along this line from $$(1, -2, 3)$$ at parameter $$t$$ is $$(1+3t, -2-mt, 3+t)$$. This point lies on the plane when $$1+3t + 2(-2-mt) - 3(3+t) + 10 = 0$$.

$$1 + 3t - 4 - 2mt - 9 - 3t + 10 = 0$$, so $$-2 - 2mt = 0$$, giving $$t = \frac{-1}{m}$$.

The distance from $$(1,-2,3)$$ to the point on the plane is $$|t|\sqrt{9 + m^2 + 1} = \frac{1}{|m|}\sqrt{m^2 + 10}$$.

Setting this equal to $$\sqrt{\frac{7}{2}}$$: $$\frac{m^2 + 10}{m^2} = \frac{7}{2}$$.

$$2(m^2 + 10) = 7m^2$$, so $$5m^2 = 20$$, giving $$m^2 = 4$$ and $$|m| = 2$$.

The answer is $$2$$.