Let $a-2b+c=1$. If $f(x) = \begin{vmatrix} x+a & x+2 & x+1 \\ x+b & x+3 & x+2 \\ x+c & x+4 & x+3 \end{vmatrix}$,then:

For non-zero,real $a, b$ and $c$,if $\left| \begin{array}{ccc} \frac{a^2+b^2}{c} & c & c \\ a & \frac{b^2+c^2}{a} & a \\ b & b & \frac{c^2+a^2}{b} \end{array} \right| = \alpha abc$,then the value of $\alpha$ is

Using properties of determinants,prove that:
$\left|\begin{array}{ccc}1 & 1+p & 1+p+q \\ 2 & 3+2 p & 4+3 p+2 q \\ 3 & 6+3 p & 10+6 p+3 q\end{array}\right|=1$

By using properties of determinants,show that:
$\left|\begin{array}{ccc}1 & x & x^{2} \\ x^{2} & 1 & x \\ x & x^{2} & 1\end{array}\right|=\left(1-x^{3}\right)^{2}$

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

If $a, b, c$ are all different from zero and $\left| \begin{array}{ccc} 1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c \end{array} \right| = 0$,then the value of $a^{-1} + b^{-1} + c^{-1}$ is