Substitute 3acosa = ccosb+bcosc;
get COSA = 1/3；
∴sinA= 2√3/3
cosB =-cos(A+C)=-cosa cosC+Sina sinC =- 1/3 cosC+2√3/3 sinC③
We also know that cosB+cosC= 2√3/3 is substituted into ③.
CosC+√2 sinC=√3，COS 2C+SIN 2C = 1。
The solution is sinC= √6/3.
It is known that A= 1.
Sine theorem: c= √3/2