What is the area of the triangle with vertices $(4, 4)$,$(3, -2)$,and $(3, -16)$?

If $px^4 + qx^3 + rx^2 + sx + t \equiv \left| \begin{array}{ccc} x^2 + 3x & x - 1 & x + 3 \\ x + 1 & 2 - x & x - 3 \\ x - 3 & x + 4 & 3x \end{array} \right|$,then $t =$

If $C = 2 \cos \theta$,then the value of the determinant $\Delta = \begin{vmatrix} C & 1 & 0 \\ 1 & C & 1 \\ 6 & 1 & C \end{vmatrix}$ is

In a square matrix $A$ of order $3$,$a_{ii}$ are the sum of the roots of the equation $x^2 - (a + b)x + ab = 0$; $a_{i, i+1}$ are the product of the roots,$a_{i, i-1}$ are all unity,and the rest of the elements are all zero. The value of the determinant of $A$ is equal to

If $\left|\begin{array}{ccc}9 & 25 & 16 \\ 16 & 36 & 25 \\ 25 & 49 & 36\end{array}\right|=K$, then $K, K+1$ are the roots of the equation

Let $0 \neq a \in \mathbb{Z}$ and $A = \begin{bmatrix} a & a & a-y \\ a & a+x & a \\ a & a & a \end{bmatrix}$ be a matrix. Then,the equation $\det(A) = 16$ represents: