Prove the following using properties of determinants: \begin{vmatrix} a+b+2c & cb \ cb & b+c+2a \ ca & c+a+2b \end{vmatrix} = 2(a+b+c)^3

LHS=\begin{vmatrix} a+b+2c & cb \ cb & b+c+2a \ ca & c+a+2b \end{vmatrix}
Applying C1 to C1+C2+C3
\begin{vmatrix} 2a+2b+2c & cb \ 2a+2b+2c & b+c+2a \ 2a+2b+2c & c+a+2b \end{vmatrix}
=(2a+2b+2c)\begin{vmatrix} 1 & cb \ 1 & b+c+2a \ 1 & c+a+2b \end{vmatrix}
Applying R2 \to R2-R1 \ R3 \to R3-R1
=(2a+2b+2c)\begin{vmatrix} 1 & cb \ 0 & b+c+a \ 0 & c+a+b \end{vmatrix}
Expanding along third row, we get
=2(a+b+c)[(b+c+a)c-bc]=2(a+b+c)[bc+c^2+ac-bc]=2(a+b+c)(ac+c^2+ab)
=2(a+b+c)[c(a+b+c)]=2(a+b+c)^2c
This is incorrect.
Let's try another approach:
LHS = \begin{vmatrix} a+b+2c & cb \ cb & b+c+2a \ ca & c+a+2b \end{vmatrix}
Applying C1 to C1+C2+C3
=\begin{vmatrix} 2(a+b+c) & cb \ 2(a+b+c) & b+c+2a \ 2(a+b+c) & c+a+2b \end{vmatrix}
=2(a+b+c)\begin{vmatrix} 1 & cb \ 1 & b+c+2a \ 1 & c+a+2b \end{vmatrix}
Applying R2 \to R2 - R1 and R3 \to R3 - R1
=2(a+b+c)\begin{vmatrix} 1 & cb \ 0 & a+b+c \ 0 & a+b+c \end{vmatrix}
Expanding along the first column
=2(a+b+c)[(a+b+c) - (a+b+c)c]=2(a+b+c)(a+b+c - cb - c^2 - ac)
=2(a+b+c)[(a+b+c)(1-c)]
This is also incorrect. There must be a mistake in the original question. Let's assume the question is correct.
Applying C1 to C1+C2+C3
=\begin{vmatrix} 2(a+b+c) & cb \ 2(a+b+c) & b+c+2a \ 2(a+b+c) & c+a+2b \end{vmatrix}=2(a+b+c) \begin{vmatrix} 1 & cb \ 1 & b+c+2a \ 1 & c+a+2b \end{vmatrix}
=2(a+b+c) \begin{vmatrix} 1 & cb \ 0 & a+b+c \ 0 & a+b+c \end{vmatrix} = 0
This shows there is an error in the problem statement.