A line with direction ratios $1,-1,2$ intersects the lines $\frac{x}{2}=\frac{y}{3}=\frac{z+1}{3}$ and $\frac{x+1}{-1}=\frac{y-2}{1}=\frac{z}{4}$ at the points P and Q , respectively. If the length of the line segment PQ is $\alpha$, then $225 \alpha^2$ is equal to:

The square of the distance of the point $(-2,-8,6)$ from the line $\frac{x-1}{1}=\frac{y-1}{2}=\frac{z}{-1}$ along the line $\frac{x+5}{1}=\frac{y+5}{-1}=\frac{z}{2}$ is equal to:

If $y=\tan ^{-1}\left(\frac{3 \cos x-4 \sin x}{4 \cos x+3 \sin x}\right)+2 \tan ^{-1}\left(\frac{x}{1+\sqrt{1-x^2}}\right)$, then $\frac{d y}{d x}$ at $x=\frac{\sqrt{3}}{2}$ is equal to :

Let $f$ be a real polynomial of degree $n$ such that $f(x)=f^{\prime}(x) f^{\prime \prime}(x)$, for all $x \in \mathbb{R}$. If $f(0)=0$, then $36\left(f^{\prime}(2)+f^{\prime \prime}(2)+\int_0^2 f(x) d x\right)$ is equal to: