$\left| {\,\begin{array}{*{20}{c}}{1/a}&{{a^2}}&{bc}\\{1/b}&{{b^2}}&{ca}\\{1/c}&{{c^2}}&{ab}\end{array}\,} \right| = $
$abc$
$1/abc$
$ab + bc + ca$
$0$
If $a \ne b \ne c,$ the value of $x$ which satisfies the equation $\left| {\,\begin{array}{*{20}{c}}0&{x - a}&{x - b}\\{x + a}&0&{x - c}\\{x + b}&{x + c}&0\end{array}\,} \right| = 0$, is
$\left| {\,\begin{array}{*{20}{c}}{19}&{17}&{15}\\9&8&7\\1&1&1\end{array}\,} \right| = $
If $n$ be the number of values of $x$ for which
matrix $\Delta (x) =\left[ {\begin{array}{*{20}{c}}
{ - x}&x&2\\
2&x&{ - x}\\
x&{ - 2}&{ - x}
\end{array}} \right]$ will be singular, then $det(\Delta\,(n))$ is
$($ where $det(B)$ denotes determinant of Matrix $B) -$
Let $\alpha \beta \gamma=45 ; \alpha, \beta, \gamma \in R$. If $x(\alpha, 1,2)+y(1, \beta, 2)$ $+z(2,3, \gamma)=(0,0,0)$ for some $x, y, z \in R, x y z \neq$ 0 , then $6 \alpha+4 \beta+\gamma$ is equal to..............