The values of $x$ in the following determinant equation,$\left| \begin{array}{ccc} a+x & a-x & a-x \\ a-x & a+x & a-x \\ a-x & a-x & a+x \end{array} \right| = 0$ are

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

If $\left| \begin{array}{ccc} x + 1 & x + 2 & x + 3 \\ x + 2 & x + 3 & x + 4 \\ x + a & x + b & x + c \end{array} \right| = 0$,then $a, b, c$ are in

If $\left|\begin{array}{ccc}1+x & 1 & 1 \\ 1+y & 1+2 y & 1 \\ 1+z & 1+z & 1+3 z\end{array}\right| = 10 k x y z \left(3+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)$,then $k = \text{ . . . . . . }$ (where $x, y, z \neq 0$ and $3+\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \neq 0$).

Let $a, b, c > 0$ and $\Delta = \begin{vmatrix} a+b & b & c \\ b+c & c & a \\ c+a & a & b \end{vmatrix}$. Then which of the following is not correct?

Evaluate the determinant: $\left|\begin{array}{ll}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{array}\right|$