If the points with position vectors $60 \hat{i}+3 \hat{j}$,$40 \hat{i}-8 \hat{j}$,and $a \hat{i}-52 \hat{j}$ are collinear,then $a$ is equal to

If $A = \begin{bmatrix} x & 1 & 2 \\ 2 & 4 & x \\ -3 & 3 & 2 \end{bmatrix}$ is a singular matrix and the distinct values of $x$ are $x_1$ and $x_2$,then $x_1 + x_2 + x_1 x_2 = $.

Find the area of the triangle with vertices at the points $(-2, -3), (3, 2), (-1, -8)$.

Let $\left|\begin{array}{ccc}x^2+x+1 & x+1 & 2x-3 \\ 3x^2-1 & x+2 & x-1 \\ x^2+5x+1 & 2x+3 & x+4\end{array}\right| = ax^4+bx^3+cx^2+dx+e$ be an identity in $x$. If $a, b, c, d$ are known,then the value of $e$ is

If $A = \begin{bmatrix} 1 & 0 & 1 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{bmatrix}$,then $\det(A)$ is equal to

Three-digit numbers $x17$,$3y6$,and $12z$,where $x, y, z$ are integers from $0$ to $9$,are divisible by a fixed constant $k$. Then the determinant $\left| \begin{array}{ccc} x & 3 & 1 \\ 7 & 6 & z \\ 1 & y & 2 \end{array} \right| + 48$ must be divisible by: