It's much easier to use $2$ (vs. $3)$ congruences, i.e. $\ 374 = (2\cdot 11)\, 17 = 22\cdot 17\ $ so
$\!\!\bmod \color{#0a0}{22}\!:\,\ \overbrace{{-}3x \equiv -15^{\phantom{.}}}^{\Large\ 19x\ \ \equiv\ \ 7_{\phantom{I}}}\! \iff x\, \equiv\, \color{#0a0}5$
$\!\!\bmod \color{#c00}{17}\!:\ \ \ \ \ 2x \equiv -10\iff x \equiv -5 \equiv \color{#0a0}{5+22}\,\color{#c00}k \equiv 5\!+\!5k\!\iff\! 5k \equiv-10\!\iff\! \color{#c00}{k \equiv -2}$
so substituting for $\,\color{#c00}k\,$ we obtain that: $\,\ x = 5 + 22(\color{#c00}{-2\!+\!17}n)\equiv \bbox[5px,border:1px solid red]{-39\equiv 335 \pmod{\!374}}$
Alternatively applying $ $ Gauss's algorithm and Inverse Reciprocity we easily compute
$\bmod 374\!:\,\ \dfrac{7}{19} \equiv \dfrac{19\cdot 7\ \ }{19\cdot 19}\equiv \dfrac{ 133}{-13} \equiv \dfrac{\color{#90f}{133+374}}{-13}\equiv -39,\ $ by $\bmod 13\!:\,\ \color{#90f}{133}\equiv 3\equiv \color{#90f}{-374}$