First, we need to understand the oxidation states of chromium in the given compounds and determine their magnetic moments based on their electronic configurations.

1. $$\mathrm{CrO}$$: In $$\mathrm{CrO}$$, the oxidation state of chromium is +2. The electronic configuration of chromium (Cr) is $$1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^1$$. In the +2 oxidation state, two electrons are removed, typically from the 4s and one of the 3d orbitals, leaving us with the configuration $$3d^4$$.

To find the spin-only magnetic moment, we use the formula:

$$\mu = \sqrt{n(n+2)} \mathrm{BM}$$

where n is the number of unpaired electrons. For $$\mathrm{Cr}^{2+}$$, we have 4 unpaired electrons in the 3d orbitals.

Thus,

$$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \mathrm{BM}$$

2. $$\mathrm{Cr}_2 \mathrm{O}_3$$: In $$\mathrm{Cr}_2 \mathrm{O}_3$$, the oxidation state of chromium is +3. The electronic configuration of $$\mathrm{Cr}^{3+}$$ is $$3d^3$$ after losing three electrons.

For $$\mathrm{Cr}^{3+}$$, there are 3 unpaired electrons.

Thus,

$$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \mathrm{BM}$$

3. $$\mathrm{CrO}_3$$: In $$\mathrm{CrO}_3$$, the oxidation state of chromium is +6. The electronic configuration of $$\mathrm{Cr}^{6+}$$ is $$3d^0$$ after losing six electrons. There are no unpaired electrons for $$\mathrm{Cr}^{6+}$$.

Since $$\mathrm{Cr}^{6+}$$ has no unpaired electrons, its spin-only magnetic moment is 0 BM.

The magnetic moment values for each compound are:

$$\mathrm{CrO}$$: 4.90 BM (basic oxide)
$$\mathrm{Cr}_2 \mathrm{O}_3$$: 3.87 BM (amphoteric oxide)
$$\mathrm{CrO}_3$$: 0 BM

The sum of the spin-only magnetic moments of the basic and amphoteric oxides is:

$$4.90 + 3.87 = 8.77 \mathrm{BM}$$

Expressing in terms of $$10^{-2} \mathrm{BM}$$,

$$8.77 \mathrm{BM} \times 100 = 877 (\times 10^{-2} \mathrm{BM})$$

Thus, the sum of spin-only magnetic moment values of basic and amphoteric oxides is approximately 877 $$\times 10^{-2} \mathrm{BM}$$ (nearest integer).