A vector $$\vec{a}$$ has components $$3p$$ and 1 with respect to a rectangular cartesian system. This system is rotated through a certain angle about the origin in the counter clockwise sense. If, with respect to new system, $$\vec{a}$$ has components $$p+1$$ and $$\sqrt{10}$$, then a value of $$p$$ is equal to:

When a coordinate system is rotated, the magnitude of a vector remains unchanged. So we use the property that $$|\vec{a}|^2$$ is the same in both coordinate systems.

In the original system, $$\vec{a}$$ has components $$3p$$ and $$1$$, so $$|\vec{a}|^2 = (3p)^2 + 1^2 = 9p^2 + 1$$.

In the rotated system, $$\vec{a}$$ has components $$(p+1)$$ and $$\sqrt{10}$$, so $$|\vec{a}|^2 = (p+1)^2 + (\sqrt{10})^2 = (p+1)^2 + 10$$.

Setting these equal: $$9p^2 + 1 = (p+1)^2 + 10$$.

Expanding the right side: $$(p+1)^2 + 10 = p^2 + 2p + 1 + 10 = p^2 + 2p + 11$$.

So we get $$9p^2 + 1 = p^2 + 2p + 11$$.

Rearranging: $$9p^2 - p^2 - 2p + 1 - 11 = 0$$, which gives $$8p^2 - 2p - 10 = 0$$.

Dividing by 2: $$4p^2 - p - 5 = 0$$.

Using the quadratic formula: $$p = \frac{1 \pm \sqrt{1 + 80}}{8} = \frac{1 \pm \sqrt{81}}{8} = \frac{1 \pm 9}{8}$$.

This gives $$p = \frac{1+9}{8} = \frac{10}{8} = \frac{5}{4}$$ or $$p = \frac{1-9}{8} = \frac{-8}{8} = -1$$.

Among the given options, $$p = -1$$ matches Option D. Therefore the answer is Option D: $$-1$$.