A body of mass $$5$$ kg moving with a uniform speed $$3\sqrt{2} \text{ m s}^{-1}$$ in $$X-Y$$ plane along the line $$y = x + 4$$. The angular momentum of the particle about the origin will be ______ kg m$$^2$$ s$$^{-1}$$.

We need to find the angular momentum of a 5 kg body moving with speed $$3\sqrt{2}$$ m/s along the line $$y = x + 4$$ about the origin.

The formula for angular momentum is $$L = mvd$$, where $$d$$ is the perpendicular distance from the point (origin) to the line of motion.

Rewriting the line $$y = x + 4$$ in standard form gives $$x - y + 4 = 0$$. Hence the perpendicular distance from $$(0, 0)$$ to this line is $$d = \frac{|0 - 0 + 4|}{\sqrt{1^2 + (-1)^2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$$ m.

Substituting into the angular momentum formula yields $$L = mvd = 5 \times 3\sqrt{2} \times 2\sqrt{2} = 5 \times 3 \times 2 \times (\sqrt{2} \times \sqrt{2}) = 5 \times 6 \times 2 = 60$$ kg m$$^2$$/s.

The answer is 60 kg m$$^2$$ s$$^{-1}$$.