Δ=a−b−c2b2c2ab−c−a2c2a2bc−a−b R1→R1+R2+R Δ=a+b+c2b2ca+b+cb−c−a2ca+b+c2bc−a−b =(a+b+c)12b2c1b−c−a2c12bc−a−b $\begin{array}{l}
C_{1} \rightarrow C_{1}-C_{y}, C_{2} \rightarrow C_{2}-C_{3} \
\Delta=(a+b+c)\left|\begin{array}{ccc}
0 & 0 & 1 \
0 & -b-c-a & 2 b \
c+a+b & c+a+b & c-a-b
\end{array}\right|
\end{array}=(a+b+c)(a+b+c)^{2}Hence,x=-2(a+b+c)$