Let $S=\left\{A=\left(\begin{array}{lll}0 & 1 & c \\ 1 & a & d \\ 1 & b & e\end{array}\right): a, b, c, d, e \in\{0,1\}\right.$ and $\left.|A| \in\{-1,1\}\right\}$, where $|A|$ denotes the determinant of $A$. Then the number of elements in $S$ is. . . . .
$10$
$15$
$16$
$20$
If $\left| {\begin{array}{*{20}{c}}{a\, + \,1}&{a\, + \,2}&{a\, + \,p}\\{a\, + \,2}&{a\, +\,3}&{a\, + \,q}\\{a\, + \,3}&{a\, + \,4}&{a\, + \,r}\end{array}} \right|$ $= 0$ , then $p, q, r$ are in :
Let the area of the triangle with vertices $A (1, \alpha)$, $B (\alpha, 0)$ and $C (0, \alpha)$ be $4\, sq.$ units. If the point $(\alpha,-\alpha),(-\alpha, \alpha)$ and $\left(\alpha^{2}, \beta\right)$ are collinear, then $\beta$ is equal to
The system of equations $\begin{array}{l}\alpha x + y + z = \alpha - 1\\x + \alpha y + z = \alpha - 1\\x + y + \alpha z = \alpha - 1\end{array}$ has no solution, if $\alpha $ is
If the system of linear equations $x-2 y+z=-4 $ ; $2 x+\alpha y+3 z=5 $ ; $3 x-y+\beta z=3$ has infinitely many solutions, then $12 \alpha+13 \beta$ is equal to