If the system of linear equation $x + 2ay + az = 0,$ $x + 3by + bz = 0,$ $x + 4cy + cz = 0$  has a non zero solution, then $a,b,c$

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

At what value of $x,$ will $\left| {\,\begin{array}{*{20}{c}}{x + {\omega ^2}}&\omega &1\\\omega &{{\omega ^2}}&{1 + x}\\1&{x + \omega }&{{\omega ^2}}\end{array}\,} \right| = 0$

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

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 :

$\left| {\,\begin{array}{*{20}{c}}{{a^2}}&{{b^2}}&{{c^2}}\\{{{(a + 1)}^2}}&{{{(b + 1)}^2}}&{{{(c + 1)}^2}}\\{{{(a - 1)}^2}}&{{{(b - 1)}^2}}&{{{(c - 1)}^2}}\end{array}\,} \right| = $