The $$x$$-intercept of a plane $$\pi$$ passing through the point $$(1,1,1)$$ is $$\frac{5}{2}$$ and the perpendicular distance from the origin to the plane $$\pi$$ is $$\frac{5}{7}$$. If the $$y$$-intercept of the plane $$\pi$$ is negative and the $$z$$-intercept is positive, then its $$y$$-intercept is

If the lines, $$\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-1}{\lambda}$$ and $$\frac{x-2}{3}=\frac{y-3}{2}=\frac{z-2}{3}$$ are coplanar, then $$\sin ^{-1}(\sin \lambda)+\cos ^{-1}(\cos \lambda)$$ is equal to

The line passing through $$(1,1,-1)$$ and parallel to the vector $$\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$$ meets the line $$\frac{x-3}{-1}=\frac{y+2}{5}=\frac{z-2}{-4}$$ at $$A$$ and the plane $$2 x-y+2 z+7=0$$ at $$B$$. Then $$A B$$ is equal to

If the vertices of the triangles are (1, 2, 3), (2, 3, 1), (3, 1, 2) and if H, G, S and I respectively denote its orthocentre, centroid, circumcentre and incentre, then H + G + S + I is equal to