If the vertices of a triangle are $(5, 2)$,$(2/3, 2)$,and $(-4, 3)$,then the area of the triangle is

Given the system of equations $a(x + y + z) = x$,$b(x + y + z) = y$,$c(x + y + z) = z$ where $a, b, c$ are non-zero real numbers. If the real numbers $x, y, z$ are such that $xyz \neq 0$,then $(a + b + c)$ is equal to-

The roots of the equation $\left| \begin{matrix} 1+x & 1 & 1 \\ 1 & 1+x & 1 \\ 1 & 1 & 1+x \end{matrix} \right| = 0$ are

If $\left| {\begin{array}{*{20}{c}}{{x_1}}&{{y_1}}&1\\{{x_2}}&{{y_2}}&1\\{{x_3}}&{{y_3}}&1\end{array}} \right| = \left| {\begin{array}{*{20}{c}}{{a_1}}&{{b_1}}&1\\{{a_2}}&{{b_2}}&1\\{{a_3}}&{{b_3}}&1\end{array}} \right|$,then the two triangles with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ and $(a_1, b_1), (a_2, b_2), (a_3, b_3)$ must be:

If $\Delta=\left|\begin{array}{ccc}x-2 & 2 x-3 & 3 x-4 \\ 2 x-3 & 3 x-4 & 4 x-5 \\ 3 x-5 & 5 x-8 & 10 x-17\end{array}\right|=Ax^{3}+Bx^{2}+Cx+D$,then $B+C$ is equal to

Let $N = \left| \begin{array}{ccc} 28 & 25 & 38 \\ 42 & 38 & 65 \\ 56 & 47 & 83 \end{array} \right|$. Then,the number of ways in which $N$ can be expressed as a product of two relatively prime divisors is: