Which one of the following does NOT equal \[\left| {\begin{array}{*{20}{c}} 1&{\text{x}}&{{{\text{x}}^2}} \\ 1&{\text{y}}&{{{\text{y}}^2}} \\ 1&{\text{z}}&{{{\text{z}}^2}} \end{array}} \right|?\] A. \[\left| {\begin{array}{*{20}{c}} 1&{{\text{x}}\left( {{\text{x}} + 1} \right)}&{{\text{x}} + 1} \\ 1&{{\text{y}}\left( {{\text{y}} + 1} \right)}&{{\text{y}} + 1} \\ 1&{{\text{z}}\left( {{\text{z}} + 1} \right)}&{{\text{z}} + 1} \end{array}} \right|\] B. \[\left| {\begin{array}{*{20}{c}} 1&{{\text{x}} + 1}&{{{\text{x}}^2} + 1} \\ 1&{{\text{y}} + 1}&{{{\text{y}}^2} + 1} \\ 1&{{\text{z}} + 1}&{{{\text{z}}^2} + 1} \end{array}} \right|\] C. \[\left| {\begin{array}{*{20}{c}} 0&{{\text{x}} – {\text{y}}}&{{{\text{x}}^2} – {{\text{y}}^2}} \\ 0&{{\text{y}} – {\text{z}}}&{{{\text{y}}^2} – {{\text{z}}^2}} \\ 1&{\text{z}}&{{{\text{z}}^2}} \end{array}} \right|\] D. \[\left| {\begin{array}{*{20}{c}} 2&{{\text{x}} + {\text{y}}}&{{{\text{x}}^2} + {{\text{y}}^2}} \\ 2&{{\text{y}} + {\text{z}}}&{{{\text{y}}^2} + {{\text{z}}^2}} \\ 1&{\text{z}}&{{{\text{z}}^2}} \end{array}} \right|\]

” option2=”\[\left| {\begin{array}{*{20}{c}} 1&{{\text{x}} + 1}&{{{\text{x}}^2} + 1} \\ 1&{{\text{y}} + 1}&{{{\text{y}}^2} + 1} \\ 1&{{\text{z}} + 1}&{{{\text{z}}^2} + 1} \end{array}} \right|\]” option3=”\[\left| {\begin{array}{*{20}{c}} 0&{{\text{x}} – {\text{y}}}&{{{\text{x}}^2} – {{\text{y}}^2}} \\ 0&{{\text{y}} – {\text{z}}}&{{{\text{y}}^2} – {{\text{z}}^2}} \\ 1&{\text{z}}&{{{\text{z}}^2}} \end{array}} \right|\]” option4=”\[\left| {\begin{array}{*{20}{c}} 2&{{\text{x}} + {\text{y}}}&{{{\text{x}}^2} + {{\text{y}}^2}} \\ 2&{{\text{y}} + {\text{z}}}&{{{\text{y}}^2} + {{\text{z}}^2}} \\ 1&{\text{z}}&{{{\text{z}}^2}} \end{array}} \right|\]” correct=”option3″]

The correct answer is $\boxed{\text{(B)}}$.

The determinant of a 3×3 matrix can be computed using the following formula:

$$\det \begin{bmatrix}
a & b & c \\
d & e & f \\
g & h & i
\end{bmatrix} = (aei – bg – ch) + (bf – ce)d + (cd – af)g$$

In this case, we have the matrix:

$$\begin{bmatrix}
1 & x & x^2 \\
1 & y & y^2 \\
1 & z & z^2
\end{bmatrix}$$

Substituting into the formula, we get:

$$\det \begin{bmatrix}
1 & x & x^2 \\
1 & y & y^2 \\
1 & z & z^2
\end{bmatrix} = (1)(z^2 – y^2) – (x)(y^2 – z^2) + (x)(z^2 – x^2)$$

Simplifying, we get:

$$\det \begin{bmatrix}
1 & x & x^2 \\
1 & y & y^2 \\
1 & z & z^2
\end{bmatrix} = x^2 – y^2 – z^2$$

This is the determinant of the matrix in option $\boxed{\text{(B)}}$.

The determinants of the matrices in options $\boxed{\text{(A)}}$, $\boxed{\text{(C)}}$, and $\boxed{\text{(D)}}$ are all equal to $0$. This is because they all have a row or column of identical elements.

In option $\boxed{\text{(A)}}$, the first row is all $1$s. This means that the determinant of the matrix is equal to $1$ times the determinant of the matrix that is left after we remove the first row and column. The determinant of the remaining matrix is equal to $0$, since it is a 2×2 matrix with a row of identical elements. Therefore, the determinant of the matrix in option $\boxed{\text{(A)}}$ is also equal to $0$.

In option $\boxed{\text{(C)}}$, the second column is all $0$s. This means that the determinant of the matrix is equal to $0$ times the determinant of the matrix that is left after we remove the second column and row. The determinant of the remaining matrix is equal to $0$, since it is a 2×2 matrix with a column of identical elements. Therefore, the determinant of the matrix in option $\boxed{\text{(C)}}$ is also equal to $0$.

In option $\boxed{\text{(D)}}$, the third row is all $1$s. This means that the determinant of the matrix is equal to $1$ times the determinant of the matrix that is left after we remove the third row and column. The determinant of the remaining matrix is equal to $0$, since it is a 2×2 matrix with a row of identical elements. Therefore, the determinant of the matrix in option $\boxed{\text{(D)}}$ is also equal to $0$.