$3^{-2} \ 3^{66} \mod 67 = 3^{-2} \mod 67 = 9^{-1} \mod 67$ (By Fermat’s Little theorem )

Using Euclidean algorithm

$67 = 7*9 + 4$ and $9 = 2*4 + 1 \Rightarrow 1 = 9 – 2*(67\ –\ 7*9) \Rightarrow 1 = 15*9\ –\ 2*67$

So, $9^{-1} \mod 67 = 15 $

Using Euclidean algorithm

$67 = 7*9 + 4$ and $9 = 2*4 + 1 \Rightarrow 1 = 9 – 2*(67\ –\ 7*9) \Rightarrow 1 = 15*9\ –\ 2*67$

So, $9^{-1} \mod 67 = 15 $