One of the roots of the given equation $\left| {\,\begin{array}{*{20}{c}}{x + a}&b&c\\b&{x + c}&a\\c&a&{x + b}\end{array}\,} \right| = 0$ 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 value of $\lambda$ and $\mu$ such that the system of equations $x+y+z=6,3 x+5 y+5 z=26, x+2 y+\lambda z=\mu$ has no solution, are :

$\left| {\begin{array}{*{20}{c}}0&a&{ - b}\\{ - a}&0&c\\b&{ - c}&0\end{array}} \right| = $

If $\omega $ be a complex cube root of unity, then $\left| {\,\begin{array}{*{20}{c}}1&\omega &{ - {\omega ^2}/2}\\1&1&1\\1&{ - 1}&0\end{array}\,} \right| = $

The roots of the equation $\left| {\,\begin{array}{*{20}{c}}{x - 1}&1&1\\1&{x - 1}&1\\1&1&{x - 1}\end{array}\,} \right| = 0$  are