Let $y=y(x)$ be the solution curve of the differential equation $\left(1+x^2\right) \mathrm{d} y+\left(y-\tan ^{-1} x\right) d x=0, y(0)=1$. Then the value of $y(1)$ is :

The number of strictly increasing functions $f$ from the set $\{1,2,3,4,5,6\}$ to the set $\{1,2,3, \ldots ., 9\}$ such that $f(i) \neq i$ for $1 \leq i \leq 6$, is equal to :

Let $(\alpha, \beta, \gamma)$ be the co-ordinates of the foot of the perpendicular drawn from the point $(5,4,2)$ on the line $\overrightarrow{\mathrm{r}}=(-\hat{i}+3 \hat{j}+\hat{k})+\lambda(2 \hat{i}+3 \hat{j}-\hat{k})$.

Then the length of the projection of the vector $\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}$ on the vector $6 \hat{i}+2 \hat{j}+3 \hat{k}$ is :

Let $a_1, a_2, a_3, \ldots$ be a G.P. of increasing positive terms such that $a_2 \cdot a_3 \cdot a_4=64$ and $a_1+a_3+a_5=\frac{813}{7}$. Then $a_3+a_5+a_7$ is equal to :