$[a\,b\,c] \ne 0$.
Now $a + 2b + 3c,\,\,\,\lambda b + 4c$ and $(2\lambda - 1)c$ will be non- coplanor, iff $(a + 2b + 3c).\{ \lambda b + 4c) \times (2\lambda - 1)c\} \ne 0$
i.e., $(a + 2b + 3c)\{ \lambda (2\lambda - 1)(b \times c)\} \ne 0$
i.e., $\lambda (2\lambda - 1)[a\,b\,c] \ne 0$, .? $\lambda \ne 0,\frac{1}{2}$
Thus given vectors will be non-coplanar for all values of $\lambda $ except two values, $\lambda = 0$ and $\lambda = 1/2$.
Trick : For coplanarity, $\left| {\,\begin{array}{*{20}{c}}1&2&3\\0&\lambda &4\\0&0&{2\lambda - 1}\end{array}\,} \right| = 0 \Rightarrow \lambda = 0,\,\frac{1}{2}$
$\therefore $ All values except two values of $\lambda $ = $0,$ $\frac{1}{2}$.
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If $I_1 = \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{3}} \, f (\tan\, \theta + \cot\, \theta )\cdot sec^2\, \theta\, d\, \theta$ &
$I_2 = \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{3}} \, f (\tan\, \theta + \cot\, \theta )\cdot cosec^2\, \theta\, d \, \theta$ ,
then the ratio $\frac{{{I_1}}}{{{I_2}}}$ :