$\left| {\begin{array}{*{20}{c}}0&a&{ - b}\\{ - a}&0&c\\b&{ - c}&0\end{array}} \right| = $

If $a \ne 6,b,c$ satisfy $\left| {\,\begin{array}{*{20}{c}}a&{2b}&{2c}\\3&b&c\\4&a&b\end{array}\,} \right| = 0,$then $abc = $

Find the area of the triangle whose vertices are $(3,8),(-4,2)$ and $(5,1)$

If ${\Delta _1} = \left| {\,\begin{array}{*{20}{c}}x&b&b\\a&x&b\\a&a&x\end{array}\,} \right|$ and ${\Delta _2} = \left| {\,\begin{array}{*{20}{c}}x&b\\a&x\end{array}\,} \right|$ are the given determinants, then

Let $A_1, A_2, A_3$ be the three A.P. with the same common difference $d$ and having their first terms as $A , A +1, A +2$, respectively. Let $a , b , c$ be the $7^{\text {th }}, 9^{\text {th }}, 17^{\text {th }}$ terms of $A_1, A_2, A_3$, respectively such that $\left|\begin{array}{lll} a & 7 & 1 \\ 2 b & 17 & 1 \\ c & 17 & 1\end{array}\right|+70=0$ If $a=29$, then the sum of first $20$ terms of an $AP$ whose first term is $c - a - b$ and common difference is $\frac{ d }{12}$, is equal to $........$.

The values of $\alpha$, for which $\left|\begin{array}{ccc}1 & \frac{3}{2} & \alpha+\frac{3}{2} \\ 1 & \frac{1}{3} & \alpha+\frac{1}{3} \\ 2 \alpha+3 & 3 \alpha+1 & 0\end{array}\right|=0$, lie in the interval