Let the line $$\frac{x}{1}=\frac{6-y}{2}=\frac{z+8}{5}$$ intersect the lines $$\frac{x-5}{4}=\frac{y-7}{3}=\frac{z+2}{1}$$ and $$\frac{x+3}{6}=\frac{3-y}{3}=\frac{z-6}{1}$$ at the points $$\mathrm{A}$$ and $$\mathrm{B}$$ respectively. Then the distance of the mid-point of the line segment $$\mathrm{AB}$$ from the plane $$2 x-2 y+z=14$$ is :

Let $$\mu$$ be the mean and $$\sigma$$ be the standard deviation of the distribution

where $$\sum f_{i}=62$$. If $$[x]$$ denotes the greatest integer $$\leq x$$, then $$\left[\mu^{2}+\sigma^{2}\right]$$ is equal to :

Let $$\vec{a}=2 \hat{i}+7 \hat{j}-\hat{k}, \vec{b}=3 \hat{i}+5 \hat{k}$$ and $$\vec{c}=\hat{i}-\hat{j}+2 \hat{k}$$. Let $$\vec{d}$$ be a vector which is perpendicular to both $$\vec{a}$$ and $$\vec{b}$$, and $$\vec{c} \cdot \vec{d}=12$$. Then $$(-\hat{i}+\hat{j}-\hat{k}) \cdot(\vec{c} \times \vec{d})$$ is equal to :

If the points $$\mathrm{P}$$ and $$\mathrm{Q}$$ are respectively the circumcenter and the orthocentre of a $$\triangle \mathrm{ABC}$$, then $$\overrightarrow{\mathrm{PA}}+\overrightarrow{\mathrm{PB}}+\overrightarrow{\mathrm{PC}}$$ is equal to :