Let $A=\left[\begin{array}{lll}\alpha & 1 & 2 \\ 2 & 3 & 0 \\ 0 & 4 & 5\end{array}\right]$ and $B=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & -5 \alpha & 0 \\ 0 & 4 \alpha & -2 \alpha\end{array}\right]+\operatorname{adj}(A)$. If $\operatorname{det}(B)=66$, then $\operatorname{det}(\operatorname{adj}(A))$ equals :

Let $\alpha=3+4+8+9+13+14+\ldots$ upto 40 terms. If $(\tan \beta)^{\frac{\alpha}{1020}}$ is a root of the equation $x^2+x-2=0, \beta \in\left(0, \frac{\pi}{2}\right)$, then $\sin ^2 \beta+3 \cos ^2 \beta$ is equal to :

A candidate has to go to the examination centre to appear in an examination. The candidate uses only one means of transportation for the entire distance out of bus, scooter and car. The probabilities of the candidate going by bus, scooter and car, respectively, are $\frac{2}{5}, \frac{1}{5}$ and $\frac{2}{5}$. The probabilities that the candidate reaches late at the examination centre are $\frac{1}{5}, \frac{1}{3}$ and $\frac{1}{4}$ if the candidate uses bus, scooter and car, respectively. Given that the candidate reached late at the examination centre, the probability that the candidate travelled by bus is :

A set of four observations has mean 1 and variance 13. Another set of six observations has mean 2 and variance 1 . Then, the variance of all these 10 observations is equal to :