Let $$y=y(x), y > 0$$, be a solution curve of the differential equation $$\left(1+x^{2}\right) \mathrm{d} y=y(x-y) \mathrm{d} x$$. If $$y(0)=1$$ and $$y(2 \sqrt{2})=\beta$$, then

Let the lines $$l_{1}: \frac{x+5}{3}=\frac{y+4}{1}=\frac{z-\alpha}{-2}$$ and $$l_{2}: 3 x+2 y+z-2=0=x-3 y+2 z-13$$ be coplanar. If the point $$\mathrm{P}(a, b, c)$$ on $$l_{1}$$ is nearest to the point $$\mathrm{Q}(-4,-3,2)$$, then $$|a|+|b|+|c|$$ is equal to

If $$\frac{1}{n+1}{ }^{n} \mathrm{C}_{n}+\frac{1}{n}{ }^{n} \mathrm{C}_{n-1}+\ldots+\frac{1}{2}{ }^{n} \mathrm{C}_{1}+{ }^{n} \mathrm{C}_{0}=\frac{1023}{10}$$ then $$n$$ is equal to :

Let $$< a_{\mathrm{n}} > $$ be a sequence such that $$a_{1}+a_{2}+\ldots+a_{n}=\frac{n^{2}+3 n}{(n+1)(n+2)}$$. If $$28 \sum_\limits{k=1}^{10} \frac{1}{a_{k}}=p_{1} p_{2} p_{3} \ldots p_{m}$$, where $$\mathrm{p}_{1}, \mathrm{p}_{2}, \ldots ., \mathrm{p}_{\mathrm{m}}$$ are the first $$\mathrm{m}$$ prime numbers, then $$\mathrm{m}$$ is equal to