Using properties of determinants prove the following: \begin{vmatrix} x & x+y & x+2y \ x & x+y & x+2y \ x & x+y & x+2y \end{vmatrix} = 9y^2(x+y)

Take L.H.S
\begin{vmatrix} x & x+y & x+2y \ x & x+y & x+2y \ x & x+y & x+2y \end{vmatrix}
Now apply C1 \rightarrow C1 + C2 + C3
\begin{vmatrix} 3x+3y & x+y & x+2y \ 3x+3y & x+y & x+2y \ 3x+3y & x+y & x+2y \end{vmatrix}
Taking common 3(x+y) from C1
3(x+y)\begin{vmatrix} 1 & x+y & x+2y \ 1 & x+y & x+2y \ 1 & x+y & x+2y \end{vmatrix}
Now apply, R1 \rightarrow R1 - R3 \ R2 \rightarrow R2 - R3
3(x+y)\begin{vmatrix} 0 & 0 & 0 \ 0 & 0 & 0 \ 1 & x+y & x+2y \end{vmatrix}
This determinant is zero because two rows are identical. Therefore, the given equation is incorrect.
Let's assume the question is:
\begin{vmatrix} x & x+y & x+2y \ x & x+y & x+2y \ x+y & x+2y & x \end{vmatrix} = 9y^2(x+y)
Applying C1 \rightarrow C1 + C2 + C3
3(x+y)\begin{vmatrix} 1 & x+y & x+2y \ 1 & x+y & x+2y \ 1 & x+2y & x \end{vmatrix}
R1 \rightarrow R1 - R3, R2 \rightarrow R2 - R3
3(x+y)\begin{vmatrix} 0 & -y & 2y \ 0 & -y & 2y \ 1 & x+2y & x \end{vmatrix} = 3(x+y)[0]
=0
Let's assume the question is:
\begin{vmatrix} x & x+y & x+2y \ x & x+y & x+2y \ x & x+y & x+2y \end{vmatrix} = 9y^2(x+y)
This determinant is equal to zero because two rows are identical. Therefore the equation is incorrect.