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

If ${a^2} + {b^2} + {c^2} = – 2$ and $f(x) = \left| {\begin{array}{*{20}{c}}{1 + {a^2}x}&{(1 + {b^2})x}&{(1 + {c^2})x}\\{(1 + {a^2})x}&{1 + {b^2}x}&{(1 + {c^2})x}\\{(1 + {a^2})x}&{(1 + {b^2})x}&{1 + {c^2}x}\end{array}} \right|$ then $f(x)$ is a polynomial of degree

If $x$ is a positive integer, then $\Delta = \left| {\,\begin{array}{*{20}{c}}{x!}&{(x + 1)!}&{(x + 2)!}\\{(x + 1)!}&{(x + 2)!}&{(x + 3)!}\\{(x + 2)!}&{(x + 3)!}&{(x + 4)!}\end{array}\,} \right|$ is equal to

The number of positive integral solutions of the equation $\left| {\begin{array}{*{20}{c}}{{x^3} + 1}&{{x^2}y}&{{x^2}z}\\{x{y^2}}&{{y^3} + 1}&{{y^2}z}\\{x{z^2}}&{y{z^2}}&{{z^3} + 1}\end{array}} \right|$ $= 11$ is

The determinant $\left| {\,\begin{array}{*{20}{c}}a&b&{a\alpha + b}\\b&c&{b\alpha + c}\\{a\alpha + b}&{b\alpha + c}&0\end{array}\,} \right| = 0$, if $a,b,c$ are in