If $\left| {\,\begin{array}{*{20}{c}}{x + 1}&3&5\\2&{x + 2}&5\\2&3&{x + 4}\end{array}\,} \right| = 0$, then $ x =$

The area of a triangle is $5$ and two of its vertices are $A(2, 1), B(3, -2)$. The third  vertex which lies on line $y = x + 3$ is-

The determinant $\left| {\,\begin{array}{*{20}{c}}a&b&{a - b}\\b&c&{b - c}\\2&1&0\end{array}\,} \right|$ is equal to zero if $a,b,c$ are in

Find values of $x$, if $\left|\begin{array}{ll}2 & 4 \\ 5 & 1\end{array}\right|=\left|\begin{array}{cc}2 x & 4 \\ 6 & x\end{array}\right|$

If $p + q + r = 0 = a + b + c$, then the value of the determinant $\left| {\,\begin{array}{*{20}{c}}{pa}&{qb}&{rc}\\{qc}&{ra}&{pb}\\{rb}&{pc}&{qa}\end{array}\,} \right|$ is

If the system of linear equations $x+y+3 z=0$

$x+3 y+k^{2} z=0$

$3 x+y+3 z=0$

has a non-zero solution $(x, y, z)$ for some $k \in R ,$ then $x +\left(\frac{ y }{ z }\right)$ is equal to