Let $\overrightarrow{a}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}},\mathrm{ }\overrightarrow{b}=-\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}, \overrightarrow{c}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}$ and $\overrightarrow{d}=\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}$. Then, the line of intersection of planes one determined by $\overrightarrow{a}, \overrightarrow{b}$ and other determined by $\overrightarrow{c}, \overrightarrow{d}$ is perpendicular to 

Equation of line of projection of the line $3x-y+2z-1=0=x+2y-z-2$ on the plane $3x+2y+z=0$ is

The orthocentre of triangle whose vertices are $A\left(a, 0, 0\right), B\left(0, b, 0\right)$ and $\mathrm{C}\left(0, 0, \mathrm{c}\right)$ is $\left(\frac{k}{a}\mathit{,}\frac{k}{b}\mathit{,}\frac{k}{c}\right)$, then $k$ is equal to

The shortest distance between the lines $2x+y+z-1=0=3x+y+2z-2$ and $x=y=z$

 If the plane $2x+3y+6z+k=$$0$ is a tangent to the sphere $x^2+y^2+z^2+2x-2y+2z-6=0,$ then a value of $k$ is