Official solution
\[\mathop {\lim }\limits_{x \to 0} {{\alpha {e^x} + \beta {e^{ - x}} + \gamma \sin x} \over {x{{\sin }^2}x}} = {2 \over 3}\]
\[ \Rightarrow \alpha + \beta = 0\] (to make indeterminant form) ...... (i)
Now,
\[\mathop {\lim }\limits_{x \to 0} {{\alpha {e^x} - \beta {e^{ - x}} + \gamma \cos x} \over {3{x^2}}} = {2 \over 3}\] (Using L-H Rule)
\[ \Rightarrow \alpha - \beta + \gamma = 0\] (to make indeterminant form) ...... (ii)
Now,
\[\mathop {\lim }\limits_{x \to 0} {{\alpha {e^x} + \beta {e^{ - x}} - \gamma \sin x} \over {6x}} = {2 \over 3}\] (Using L-H Rule)
\[ \Rightarrow {{\alpha - \beta + \gamma } \over 6} = {2 \over 3}\]
\[ \Rightarrow \alpha - \beta + \gamma = 4\] ...... (iii)
\[ \Rightarrow \gamma = - 2\]
and (i) + (ii)
\[2\alpha = - \gamma \]
\[ \Rightarrow \alpha = 1\] and \[\beta = - 1\]
and \[\alpha {\beta ^2} + \beta {\gamma ^2} + \gamma {\alpha ^2} + 3 = 1 - 4 - 2 + 3 = - 2\]