The product of a matrix and its transpose is an identity matrix. The value of the determinant of this matrix is:

Prove that $\left|\begin{array}{ccc}a & a+b & a+b+c \\ 2a & 3a+2b & 4a+3b+2c \\ 3a & 6a+3b & 10a+6b+3c\end{array}\right|=a^{3}$

If $x, y, z$ are all positive and are the $p$-th, $q$-th, and $r$-th terms of a geometric progression respectively, then the value of the determinant $\left|\begin{array}{lll} \log x & p & 1 \\ \log y & q & 1 \\ \log z & r & 1 \end{array}\right|$ equals:

If $a - 2b + c = 1$,then the value of $\left| \begin{array}{ccc} x + 1 & x + 2 & x + a \\ x + 2 & x + 3 & x + b \\ x + 3 & x + 4 & x + c \end{array} \right|$ is

If $k \in R$ and $\operatorname{det} A = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} = K$,then $\operatorname{det} B = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 + 2a_1 & b_2 + 2b_1 & c_2 + 2c_1 \\ a_3 & b_3 & c_3 \end{vmatrix}$ is equal to

If $\Delta=\left|\begin{array}{lll}1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2\end{array}\right|$ and $\Delta_1=\left|\begin{array}{ccc}1 & 1 & 1 \\ b c & c a & a b \\ a & b & c\end{array}\right|$,then