The solutions of the equation $\left| \begin{array}{ccc} 1 & 1 & x \\ p+1 & p+1 & p+x \\ 3 & x+1 & x+2 \end{array} \right| = 0$ are:

If $\left|\begin{array}{ccc}2 a & x_{1} & y_{1} \\ 2 b & x_{2} & y_{2} \\ 2 c & x_{3} & y_{3}\end{array}\right|=\frac{a b c}{2} \neq 0$,then the area of the triangle whose vertices are $\left(\frac{x_{1}}{a}, \frac{y_{1}}{a}\right), \left(\frac{x_{2}}{b}, \frac{y_{2}}{b}\right), \left(\frac{x_{3}}{c}, \frac{y_{3}}{c}\right)$ 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-

If $A = \begin{bmatrix} 5 & 5x & x \\ 0 & x & 5x \\ 0 & 0 & 5 \end{bmatrix}$ and $|A^2| = 25$,then $|x|$ is equal to

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

$\left|\begin{array}{ccc}\frac{-bc}{a^2} & \frac{c}{a} & \frac{b}{a} \\ \frac{c}{b} & \frac{-ac}{b^2} & \frac{a}{b} \\ \frac{b}{c} & \frac{a}{c} & \frac{-ab}{c^2}\end{array}\right| = $