The number of solutions of the equations $x + 4y – z = 0,$ $3x – 4y – z = 0,\,x – 3y + z = 0$ 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

If $\left| {\,\begin{array}{*{20}{c}}{{x^2} + x}&{x + 1}&{x – 2}\\{2{x^2} + 3x – 1}&{3x}&{3x – 3}\\{{x^2} + 2x + 3}&{2x – 1}&{2x – 1}\end{array}\,} \right| = Ax – 12$, then the value of $A $ is

If $a \ne p,b \ne q,c \ne r$ and $\left| {\,\begin{array}{*{20}{c}}p&b&c\\{p + a}&{q + b}&{2c}\\a&b&r\end{array}\,} \right|$ =$ 0$, then $\frac{p}{{p – a}} + \frac{q}{{q – b}} + \frac{r}{{r – c}} = $

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-