If lim_x → 0 (e^( a -1) x +2 cos b x+( c -2) e^-x)/(x cos x- log_e (1+x))=2, then a^2+ b^2+ c^2 is equal to:

If $\displaystyle \lim _{x \rightarrow 0} \frac{\mathrm{e}^{(\mathrm{a}-1) x}+2 \cos \mathrm{~b} x+(\mathrm{c}-2) \mathrm{e}^{-x}}{x \cos x-\log _{\mathrm{e}}(1+x)}=2$, then $\displaystyle \mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2$ is equal to :