$f(x) = \left| \begin{array}{ccc} 1 & x & x+1 \\ 2x & x(x-1) & (x+1)x \\ 3x(x-1) & x(x-1)(x-2) & (x+1)x(x-1) \end{array} \right|$, then $f(100)$ is equal to:

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

Using the property of determinants and without expanding,prove that:
$\left|\begin{array}{lll}b+c & q+r & y+z \\ c+a & r+p & z+x \\ a+b & p+q & x+y\end{array}\right|=2\left|\begin{array}{lll}a & p & x \\ b & q & y \\ c & r & z\end{array}\right|$

Prove that $\Delta = \left| \begin{array}{ccc} a+bx & c+dx & p+qx \\ ax+b & cx+d & px+q \\ u & v & w \end{array} \right| = (1-x^2) \left| \begin{array}{ccc} a & c & p \\ b & d & q \\ u & v & w \end{array} \right|$

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

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 $\det(-A) = (-1)^n \det(A)$.