$y=3e^{x}=> y=\large\frac{a}{3}$$e^{-x}$

$\large\frac{dy}{dx} $$=3e^x$

$\large\frac{dy}{dx}=\frac{a}{3} $$ \times e^{-x} \times -1$

$m_1=3e^x$

$m_2=-\large\frac{a}{3} e^{-x}$

$m_1m_2=-1$

$3e^x \times \large\frac{-a}{3} $$e^{-x}=-1$

$-a=-1$

$a=1$

Hence 2 is the correct answer.