$\left| {\begin{array}{*{20}{c}}{x + 1}&{x + 2}&{x + 4}\\{x + 3}&{x + 5}&{x + 8}\\{x + 7}&{x + 10}&{x + 14}\end{array}} \right| = $

The value of the determinant $\left| \begin{array}{ccc} a & a+b & a+2b \\ a+2b & a & a+b \\ a+b & a+2b & a \end{array} \right|$ is

Statement-$1$: The determinant of a skew-symmetric matrix of order $3$ is zero.
Statement-$2$: For any square matrix $A$ of order $n$,$\det(A^T) = \det(A)$ and $\det(-A) = (-1)^n \det(A)$.

Evaluate $\Delta=\left|\begin{array}{lll}3 & 2 & 3 \\ 2 & 2 & 3 \\ 3 & 2 & 3\end{array}\right|$

Using properties of determinants,prove that:
$\left|\begin{array}{ccc}\alpha & \alpha^{2} & \beta+\gamma \\ \beta & \beta^{2} & \gamma+\alpha \\ \gamma & \gamma^{2} & \alpha+\beta\end{array}\right|=(\beta-\gamma)(\gamma-\alpha)(\alpha-\beta)(\alpha+\beta+\gamma)$

If the determinant $\left| \begin{array}{ccc} a+p & 1+x & u+f \\ b+q & m+y & v+g \\ c+r & n+z & w+h \end{array} \right|$ splits into exactly $K$ determinants of order $3$,each element of which contains only one term,then the value of $K$ is: