Using properties of determinants, prove that \begin{vmatrix} 1 & 1 & 1+3x \ 1 & 1 & 1+3y \ 1 & 1 & 1+3z \end{vmatrix} = 9(3xyz+xy+yz+zx)

Given Matrix is \begin{vmatrix} 1 & 1 & 1+3x \ 1 & 1 & 1+3y \ 1 & 1 & 1+3z \end{vmatrix}
Applying R2 \rightarrow R2 - R1, we have \begin{vmatrix} 1 & 1 & 1+3x \ 0 & 0 & 3y-3x \ 1 & 1 & 1+3z \end{vmatrix}
Applying R3 \rightarrow R3 - R1, we have \begin{vmatrix} 1 & 1 & 1+3x \ 0 & 0 & 3(y-x) \ 0 & 0 & 3(z-x) \end{vmatrix}
Now, Solving it as Determinant from C1 = 1(3(y-x))(3(z-x)) = 9(y-x)(z-x) = 9(yz - xy - xz + x^2) = 9yz - 9xy - 9xz + 9x^2
This is not equal to 9(3xyz + xy + yz + zx). There must be a mistake in the question or the solution provided. Let's try another approach.
Let's apply the given operations to the original determinant:
\begin{vmatrix} 1 & 1 & 1+3x \ 1 & 1 & 1+3y \ 1 & 1 & 1+3z \end{vmatrix}
Applying R2 → R2 - R1 and R3 → R3 - R1:
\begin{vmatrix} 1 & 1 & 1+3x \ 0 & 0 & 3y-3x \ 0 & 0 & 3z-3x \end{vmatrix} = 0
The determinant is 0 because the second and third rows are linearly dependent. Therefore, the given equation is incorrect. The determinant is actually 0, not 9(3xyz + xy + yz + zx).