The number of real values of $\lambda $ for which the system of linear equations $2x + 4y - \lambda z = 0$ ;$4x + \lambda y + 2z = 0$ ; $\lambda x + 2y+ 2z = 0$ has infinitely many solutions, is
$0$
$1$
$2$
$3$
If $a$, $b$, $c$, $d$, $e$, $f$ are in $G.P$., then the value of $\left| {\begin{array}{*{20}{c}}
{{a^2}}&{{d^2}}&x \\
{{b^2}}&{{e^2}}&y \\
{{c^2}}&{{f^2}}&z
\end{array}} \right|$ depends on
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..............
Let $[.]$ , $ \{.\} $ and $sgn$$(.)$ denotes greatest integer function, fractional part function and signum function respectively, then value of determinant
$\left| {\begin{array}{*{20}{c}}
{\left[ \pi \right]}&{amp(1 + i\sqrt 3 )}&1 \\
1&0&2 \\
{\operatorname{sgn} ({{\cot }^{ - 1}}x)}&1&{\{ \pi \} }
\end{array}} \right|$ is-
Statement $-1$ : The system of linear equations
$x + \left( {\sin \,\alpha } \right)y + \left( {\cos \,\alpha } \right)z = 0$
$x + \left( {\cos \,\alpha } \right)y + \left( {\sin \alpha } \right)z = 0$
$x - \left( {\sin \,\alpha } \right)y - \left( {\cos \alpha } \right)z = 0$
has a non-trivial solution for only one value of $\alpha $ lying in the interval $\left( {0\,,\,\frac{\pi }{2}} \right)$
Statement $-2$ : The equation in $\alpha $
$\left| {\begin{array}{*{20}{c}}
{\cos {\mkern 1mu} \alpha }&{\sin {\mkern 1mu} \alpha }&{\cos {\mkern 1mu} \alpha } \\
{\sin {\mkern 1mu} \alpha }&{\cos {\mkern 1mu} \alpha }&{\sin {\mkern 1mu} \alpha } \\
{\cos {\mkern 1mu} \alpha }&{ - \sin {\mkern 1mu} \alpha }&{ - \cos {\mkern 1mu} \alpha }
\end{array}} \right| = 0$
has only one solution lying in the interval $\left( {0\,,\,\frac{\pi }{2}} \right)$
The values of $x $ in the following determinant equation, $\left| {\,\begin{array}{*{20}{c}}{a + x}&{a - x}&{a - x}\\{a - x}&{a + x}&{a - x}\\{a - x}&{a - x}&{a + x}\end{array}\,} \right| = 0$ are