The given system of linear equation is

ax + 2y = $$\lambda$$

3x $$-$$ 2y = $$\mu$$

By Cramer's rule, we have

$$\Delta = \left| {\matrix{ a & 2 \cr 3 & { - 2} \cr } } \right| = - 2a - 6 = - 2(a + 3)$$

$${\Delta _1} = \left| {\matrix{ \lambda & 2 \cr \mu & { - 2} \cr } } \right| = - 2\lambda - 2\mu = - 2(\lambda + \mu )$$

$${\Delta _2} = \left| {\matrix{ a & \lambda \cr 3 & \mu \cr } } \right| = (a\mu - 3\lambda )$$

For unique solution: $$\Delta$$ $$\ne$$ 0 $$\Rightarrow$$ a + 3 $$\ne$$ 0 $$\Rightarrow$$ a $$\ne$$ $$-$$3, where $$\lambda$$ and $$\mu$$ can take any values.

Hence, option (B) is correct.

For infinite solution: $$\Delta$$ = 0, $$\Delta$$₁ and $$\Delta$$₂ are zero.

That is, a = $$-$$3 and $$\lambda$$ + $$\mu$$ = 0 or a$$\mu$$ $$-$$ 3$$\lambda$$ = 0.

Hence, options (C) are correct.

For no solution: $$\Delta$$ = 0 and either $$\Delta$$₁ or $$\Delta$$₂ is non-zero.

That is, a = $$-$$3 and $$\lambda$$ + $$\mu$$ $$\ne$$ 0.

Hence, option (D) is correct.