$\left| \begin{array}{ccc} a - b & b - c & c - a \\ x - y & y - z & z - x \\ p - q & q - r & r - p \end{array} \right| = $

If $A=\left|\begin{array}{ccc}a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3}\end{array}\right|$ and $B=\left|\begin{array}{ccc}c_{1} & c_{2} & c_{3} \\ a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3}\end{array}\right|$,then

If ${I_1} = \int\limits_1^{\sin \theta } {\frac{x}{{1 + x^2}}} \,dx$ and ${I_2} = \int\limits_1^{\csc \theta } {\frac{{dx}}{{x\left( {{x^2} + 1} \right)}}}$; then the value of $\left| {\begin{array}{*{20}{c}} {{I_1}}&{I_1^2}&{{I_2}} \\ {{e^{{I_1} + {I_2}}}}&{I_2^2}&{ - 1} \\ 1&{I_1^2 + I_2^2}&{ - 1} \end{array}} \right|$ is

By using properties of determinants,show that:
$\left|\begin{array}{lll}1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2}\end{array}\right|=(a-b)(b-c)(c-a)$

If $\Delta = \begin{vmatrix} x+y+z^2 & x^2+y+z & x+y^2+z \\ z^2 & x^2 & y^2 \\ x+y & y+z & x+z \end{vmatrix}$,(where $x \neq y \neq z$ and $x, y, z \in \mathbb{R} - \{0\}$),then $\Delta = $ . . . . . . .

Using the property of determinants and without expanding,prove that $\left|\begin{array}{lll}x & a & x+a \\ y & b & y+b \\ z & c & z+c\end{array}\right|=0$.