If $x, y, z$ are in arithmetic progression with common difference $d$,$x \neq 3d$,and the determinant of the matrix $\begin{bmatrix} 3 & 4\sqrt{2} & x \\ 4 & 5\sqrt{2} & y \\ 5 & k & z \end{bmatrix}$ is zero,then the value of $k^2$ is ..... .

Let $A=\begin{bmatrix} 2 & 2+p & 2+p+q \\ 4 & 6+2p & 8+3p+2q \\ 6 & 12+3p & 20+6p+3q \end{bmatrix}$. If $\operatorname{det}(\operatorname{adj}(\operatorname{adj}(3A)))=2^m \cdot 3^n$,where $m, n \in N$,then $m+n$ is equal to:

One of the roots of the given equation $\left| \begin{array}{ccc} x+a & b & c \\ b & x+c & a \\ c & a & x+b \end{array} \right| = 0$ is

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?

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