Show that

$\Delta=\left|\begin{array}{ccc}
(y+z)^{2} & x y & z x \\
x y & (x+z)^{2} & y z \\
x z & y z & (x+y)^{2}
\end{array}\right|=2 x y z(x+y+z)^{3}$

Using the property of determinants and without expanding, Prove that 

$\left|\begin{array}{lll}x & a & x+a \\ y & b & y+b \\ z & c & z+c\end{array}\right|=0$

The value of $\theta$ lying between $\theta = 0$ and $\theta = \pi /2$ and satisfying the equation : $\left| {\,\begin{array}{*{20}{c}} {1\,\, + \,\,{{\sin }^2}\,\theta }&{{{\cos }^2}\,\theta }&{4\,\sin \,4\,\theta }\\ {{{\sin }^2}\,\theta }&{1\,\, + \,\,{{\cos }^2}\,\theta }&{4\,\sin \,4\,\theta }\\ {{{\sin }^2}\,\theta }&{{{\cos }^2}\,\theta }&{1\,\, + \,\,4\,\sin \,4\,\theta } \end{array}\,} \right|$ $= 0$ are :

The value of $\left| {\,\begin{array}{*{20}{c}}a&{a + b}&{a + 2b}\\{a + 2b}&a&{a + b}\\{a + b}&{a + 2b}&a\end{array}\,} \right|$ is equal to

$\left| {\,\begin{array}{*{20}{c}}1&{1 + ac}&{1 + bc}\\1&{1 + ad}&{1 + bd}\\1&{1 + ae}&{1 + be}\end{array}\,} \right| = $