$\left| {\begin{array}{*{20}{c}}
{4 + {x^2}}&{ – 6}&{ – 2}\\
{ – 6}&{9 + {x^2}}&3\\
{ – 2}&3&{1 + {x^2}}
\end{array}} \right|$ $;(x\neq0)$ is not divisible by

The sum of the real roots of the equation $\left| {\begin{array}{*{20}{c}}
x&{ – 6}&{ – 1}\\
2&{ – 3x}&{x – 3}\\
{ – 3}&{2x}&{x = 2}
\end{array}} \right| = 0$ is equal to

If the system of linear equations $x+ ay+z\,= 3$ ; $x + 2y+ 2z\, = 6$ ; $x+5y+ 3z\, = b$ has no solution, then

If the system of equations $ax + y + z = 0 , x + by + z = 0 \, \& \, x + y + cz = 0$ $(a, b, c \ne 1)$ has a non-trivial solution, then the value of $\frac{1}{{1\, – \,a}}\,\, + \,\,\frac{1}{{1\, – \,b}}\,\, + \,\,\frac{1}{{1\, – \,c}}$ is :

If $\left| {\,\begin{array}{*{20}{c}}a&b&c\\m&n&p\\x&y&z\end{array}\,} \right| = k$, then $\left| {\,\begin{array}{*{20}{c}}{6a}&{2b}&{2c}\\{3m}&n&p\\{3x}&y&z\end{array}\,} \right| = $