MCQ
If the points whose position, vectors are $3i - 2j - k,$ $2i + 3j - 4k,$ $ - i + j + 2k$and $4i + 5j + \lambda k$ lie on a plane, then $\lambda = $
- ✓$ - \frac{{146}}{{17}}$
- B$\frac{{146}}{{17}}$
- C$ - \frac{{17}}{{146}}$
- D$\frac{{17}}{{146}}$
Since the points are coplanar,
So, $[d\,b\,c] + [d\,c\,a] + [d\,a\,b] = [a\,b\,c]$
$ \Rightarrow \left| {\begin{array}{*{20}{c}}4&5&\lambda \\2&3&{ - 4}\\{ - 1}&1&2\end{array}} \right| + \left| {\begin{array}{*{20}{c}}4&5&\lambda \\{ - 1}&1&2\\3&{ - 2}&{ - 1}\end{array}} \right| + \left| {\,\begin{array}{*{20}{c}}4&5&\lambda \\3&{ - 2}&{ - 1}\\2&3&{ - 4}\end{array}\,} \right|$
$ = \left| {\begin{array}{*{20}{c}}3&{ - 2}&{ - 1}\\2&3&{ - 4}\\{ - 1}&1&2\end{array}} \right|$
$ \Rightarrow 40 + 5\lambda + 37 - \lambda + 94 + 13\lambda = 25 \Rightarrow \lambda = \frac{{ - 146}}{{17}}.$
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$f(x)=\left\{\begin{array}{ll} \frac{\cos ^{-1}\left(1-\{x\}^{2}\right) \sin ^{-1}(1-\{x\})}{\{x\}-\{x\}^{3}}, & x \neq 0 \\ \alpha, & x=0 \end{array}\right.$
is continuous at $x=0,$ where $\{x\}=x-[x],[x]$ is the greatest integer less than or equal to $X$.
Then :