Tọa độ \(A\) thỏa \(\left\{\begin{array}{l}\frac{x-3}{1}=\frac{y-4}{1}=\frac{z+8}{-4} \\ x+z-1=0\end{array} \Rightarrow A(1 ; 2 ; 0)\right.\).

\(B(3+t ; 4+t ;-8-4 t)\) với \(t>-3\)

\(A B=3 \sqrt{2} \Leftrightarrow(t+2)^{2}+(t+2)^{2}+16(t+2)^{2}=18 \Rightarrow t=-1\) nên \(B(2;3;-4)\)

\(A C=A B \sin 60^{\circ}=\frac{3 \sqrt{6}}{2} ; B C=A B \cdot \cos 60^{\circ}=\frac{3 \sqrt{2}}{2} \text {. }\)

Ta có hệ \(\left\{\begin{array}{l}a+c=1 \\ (a-1)^{2}+(b-2)^{2}+c^{2}=\frac{27}{2} \\ (a-2)^{2}+(b-3)^{2}+(c+4)^{2}=\frac{9}{2}\end{array} \Leftrightarrow\left\{\begin{array}{l}a=\frac{7}{2} \\ b=3 \\ c=-\frac{5}{2}\end{array}\right.\right.\)

\(C\left(\frac{7}{2} ; 3 ;-\frac{5}{2}\right)\)