I assume [tex]T\sim\mathcal N(\mu,\sigma^2)[/tex] refers to a random variable [tex]T[/tex] following a [tex]\mathcal N[/tex]ormal distribution with mean [tex]\mu[/tex] and variance [tex]\sigma^2[/tex].
Recall that the moment generating function for a random variable [tex]T\sim\mathcal N(\mu,\sigma^2)[/tex] is
[tex]M_T(t)=\exp\left(\mu t+\dfrac12\sigma^2t^2\right)[/tex]
and that the moment generating function for a linear combination of random variables is equivalent to the product of each individual random variables moment generating functions; that is, if [tex]T=\displaystyle\sum_ia_iT_i[/tex], then
[tex]M_T(t)=\displaystyle\prod_iM_{T_i}(a_it)[/tex]
So you have
[tex]M_S(t)=M_{3X-2Y+Z}(t)=M_X(3t)\times M_Y(-2t)\times M_Z(t)[/tex]
with
[tex]M_X(3t)=\exp\left(3(3t)+\dfrac12(4)(3t)^2\right)=\exp\left(9t+\dfrac12(36)t^2\right)[/tex]
[tex]M_Y(-2t)=\exp\left(-2(-2t)+\dfrac12(6)(-2t)^2\right)=\exp\left(4t+\dfrac12(24)t^2\right)[/tex]
[tex]M_Z(t)=\exp\left(t+\dfrac12(3)t^2\right)[/tex]
which means
[tex]M_S(t)=\exp\left((9+4+1)t+\dfrac12(36+24+3)t^2\right)=\exp\left(14t+\dfrac12(63)t^2\right)[/tex]
This is the moment generating function of yet another normal distribution, so that [tex]S\sim\mathcal N(14,63)[/tex], with expectation 14 and variance 63.
