The equation of line passing through the point $$(1,2,3)$$ and perpendicular to the lines $$\frac{x-2}{3}=\frac{y-1}{2}=\frac{z+1}{-2}$$ and $$\frac{x}{2}=\frac{y}{-3}=\frac{z}{1}$$ is

The angle between the line $$\frac{x+1}{2}=\frac{y-2}{1}=\frac{z-3}{-2}$$ and plane $$x-2 y-\lambda z=3$$ is $$\cos ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)$$, then value of $$\lambda$$ is

If the direction cosines $$l, \mathrm{~m}, \mathrm{n}$$ of two lines are connected by relations $$l-5 \mathrm{~m}+3 \mathrm{n}=0$$ and $$7 l^2+5 \mathrm{~m}^2-3 \mathrm{n}^2=0$$, then value of $$l+\mathrm{m}+\mathrm{n}$$ is

The mirror image of the point $$(1,2,3)$$ in a plane is $$\left(-\frac{7}{3},-\frac{4}{3},-\frac{1}{3}\right)$$. Thus, the point _________ lies on this plane.