Let $P$ be a matrix of order $3 \times 3$ such that all the entries in $P$ are from the set $\{-1, 0, 1\}$. Then,the maximum possible value of the determinant of $P$ is:

If $a_{1}, a_{2}, a_{3}, \ldots, a_{9}$ are in $AP$,then the value of $\left|\begin{array}{lll}a_{1} & a_{2} & a_{3} \\ a_{4} & a_{5} & a_{6} \\ a_{7} & a_{8} & a_{9}\end{array}\right|$ is

If $\left| \begin{array}{ccc} a & b & a\alpha - b \\ b & c & b\alpha - c \\ 2 & 1 & 0 \end{array} \right| = 0$ and $\alpha \neq \frac{1}{2}$,then

If $a_i^2 + b_i^2 + c_i^2 = 1$ for $(i = 1, 2, 3)$ and $a_i a_j + b_i b_j + c_i c_j = 0$ for $(i \ne j, i, j = 1, 2, 3)$,then the value of $\left| \begin{array}{ccc} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{array} \right|^2$ is

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

Let $S$ be the set of all values of $\theta \in [-\pi, \pi]$ for which the system of linear equations
$x + y + \sqrt{3} z = 0$
$-x + (\tan \theta) y + \sqrt{7} z = 0$
$x + y + (\tan \theta) z = 0$
has a non-trivial solution. Then $\frac{120}{\pi} \sum_{\theta \in S} \theta$ is equal to