If $$\lim_{x \rightarrow 0} \frac{e^{(a-1)x}+2\cos bx+(c-2)e^{-x}}{x \cos x-\log_{e}{(1+x)}} =2$$, then $$a^{2}+b^{2}+c^{2}$$ is equal to :

Expand the denominator:

Using Taylor series: $$x(1 - \frac{x^2}{2} + \dots) - (x - \frac{x^2}{2} + \frac{x^3}{3} \dots) = \frac{x^2}{2} + O(x^3)$$.

Analyze the numerator:

For the limit to be finite, the numerator and its first derivative must be zero at $$x=0$$ (since the denominator is $$O(x^2)$$).

o At $$x=0$$: $$e^0 + 2\cos(0) + (c-2)e^0 = 1 + 2 + c - 2 = c + 1 = 0 \implies \mathbf{c = -1}$$.

o First Derivative at $$x=0$$: $$(a-1)e^0 - 2b\sin(0) - (c-2)e^0 = (a-1) - (-1-2) = a - 1 + 3 = a + 2 = 0 \implies \mathbf{a = -2}$$.

Second Derivative for the limit value:

Using L'Hôpital's rule twice or series expansion, the limit is $$\frac{Num''(0)}{Den''(0)} = 2$$.

$$Den''(0) = 1$$.

$$Num''(x) = (a-1)^2 e^{(a-1)x} - 2b^2 \cos bx + (c-2)e^{-x}$$.

At $$x=0$$: $$(-2-1)^2 - 2b^2 + (-1-2) = 9 - 2b^2 - 3 = 6 - 2b^2$$.

Set $$\frac{6 - 2b^2}{1} = 2 \implies 2b^2 = 4 \implies \mathbf{b^2 = 2}$$.

$$a^2 + b^2 + c^2 = (-2)^2 + 2 + (-1)^2 = 4 + 2 + 1 = \mathbf{7}$$.

Correct Option: B