Find equation of line joining $(3,1)$ and $(9,3)$ using determinants
$x-3 y=2$
$x-3 y=0$
$x+3 y=0$
$x-3 y=10$
If $n \ne 3k$ and 1, $\omega ,{\omega ^2}$ are the cube roots of unity, then $\Delta = \left| {\,\begin{array}{*{20}{c}}1&{{\omega ^n}}&{{\omega ^{2n}}}\\{{\omega ^{2n}}}&1&{{\omega ^n}}\\{{\omega ^n}}&{{\omega ^{2n}}}&1\end{array}\,} \right|$ has the value
Let $\alpha $ and $\beta $ be the roots of the equation $x^2 + x + 1 = 0.$ Then for $y \ne 0$ in $R,$ $\left| {\begin{array}{*{20}{c}}
{y\, + \,1}&\alpha &\beta \\
\alpha &{y\, + \,\beta }&1\\
\beta &1&{y\, + \,\alpha }
\end{array}} \right|$ is equal to
Find the equation of the line joining $\mathrm{A}(1,3)$ and $\mathrm{B}(0,0)$ using determinants and find $\mathrm{k}$ if $\mathrm{D}(\mathrm{k}, 0)$ is a point such that area of triangle $\mathrm{ABD}$ is $3 \,\mathrm{sq}$ $\mathrm{units}$.
$\left| {\begin{array}{*{20}{c}}
{4 + {x^2}}&{ - 6}&{ - 2}\\
{ - 6}&{9 + {x^2}}&3\\
{ - 2}&3&{1 + {x^2}}
\end{array}} \right|$ $;(x\neq0)$ is not divisible by
A root of the equation $\left| {\,\begin{array}{*{20}{c}}{3 - x}&{ - 6}&3\\{ - 6}&{3 - x}&3\\3&3&{ - 6 - x}\end{array}\,} \right| = 0$ is