Using the law of sines:
[tex]\begin{gathered} \frac{AB}{\sin(C)}=\frac{AC}{\sin (B)} \\ so\colon \\ \sin (C)=\frac{AB\cdot\sin (B)}{AC} \\ \sin (C)=\frac{12\cdot\sin (67)}{15} \\ C=\sin ^{-1}(\frac{12\cdot\sin(67)}{15}) \\ C\approx47.43^{\circ} \end{gathered}[/tex]Using the triangle sum theorem:
[tex]\begin{gathered} m\angle A+m\angle B+m\angle C=180 \\ so\colon \\ x+47.43+67=180 \\ x=180-67-47.43 \\ x=m\angle C=m\angle CAB=65.57^{\circ} \end{gathered}[/tex]
