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