Let a triangle PQR be such that P and Q lie on the line $\frac{x+3}{8}=\frac{y-4}{2}=\frac{z+1}{2}$ and are at a distance of 6 units from $R(1,2,3)$. If $(\alpha, \beta, \gamma)$ is the centroid of $\Delta P Q R$, then $\alpha+\beta+\gamma$ is equal to :

If the distance of the point $(a, 2,5)$ from the image of the point $(1,2,7)$ in the line $\frac{x}{1}=\frac{y-1}{1}=\frac{z-2}{2}$ is 4 , then the sum of all possible values of $a$ is equal to :

Let $O$ be the origin, $\overrightarrow{O P}=\vec{a}$ and $\overrightarrow{O Q}=\vec{b}$. If $R$ is the point on $\overrightarrow{O P}$ such that $\overrightarrow{O P}=5 \overrightarrow{O R}$, and $M$ is the point such that $\overrightarrow{O Q}=5 \overrightarrow{R M}$, then $\overrightarrow{P M}$ is equal to :

Let $f(x)=\lim \limits_{y \rightarrow 0} \frac{(1-\cos (x y)) \tan (x y)}{y^3}$. Then the number of solutions of the equation $f(x)=\sin x$, $x \in \mathbf{R}$ is :