$$X$$ intercept of the plane containing the line of intersection of the planes $$x-2 y+z+2=0$$ and $$3 x-y-z+1=0$$ and also passing through $$(1,1,1)$$ is

Let $$L_1$$ (resp, $$L_2$$ ) be the line passing through $$2 \hat{\mathbf{i}}-\hat{\mathbf{k}}$$ (resp. $$2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-3 \hat{\mathbf{k}})$$ and parallel to $$3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$$ ( resp. $$\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$$ ). Then the shortest distance between the lines $$L_1$$ and $$L_2$$ is equal to

If the points (2, 4, $$-$$1), (3, 6, $$-$$1) and (4, 5, $$-$$1) are three consecutive vertices of a parallelogram, then its fourth vertex is

$$A(-1,2-3), B(5,0,-6)$$ and $$C(0,4,-1)$$ are the vertices of a $$\triangle A B C$$. The direction cosines of internal bisector of $$\angle B A C$$ are