If the lines $x + 2ay + a = 0, x + 3by + b = 0$ and $x + 4cy + c = 0$ are concurrent, then $a, b$ and $c$ are in :-
Solution of the equation $\left| {\,\begin{array}{*{20}{c}}1&1&x\\{p + 1}&{p + 1}&{p + x}\\3&{x + 1}&{x + 2}\end{array}\,} \right| = 0$ are
If $\left| {\,\begin{array}{*{20}{c}}a&b&c\\m&n&p\\x&y&z\end{array}\,} \right| = k$, then $\left| {\,\begin{array}{*{20}{c}}{6a}&{2b}&{2c}\\{3m}&n&p\\{3x}&y&z\end{array}\,} \right| = $
If $A\, = \,\left[ \begin{gathered}
1\ \ \ \,1\ \ \ \,2\ \ \ \hfill \\
0\ \ \ \,2\ \ \ \,1\ \ \ \hfill \\
1\ \ \ \,0\ \ \ \,2\ \ \ \hfill \\
\end{gathered} \right]$ and $A^3 = (aA-I) (bA-I)$,where $a, b$ are integers and $I$ is a $3 × 3$ unit matrix then value of $(a + b)$ is equal to
Let $\alpha \beta \gamma=45 ; \alpha, \beta, \gamma \in R$. If $x(\alpha, 1,2)+y(1, \beta, 2)$ $+z(2,3, \gamma)=(0,0,0)$ for some $x, y, z \in R, x y z \neq$ 0 , then $6 \alpha+4 \beta+\gamma$ is equal to..............