The shortest distance between the lines

$${{x - 1} \over 0} = {{y + 1} \over { - 1}} = {z \over 1}$$

and x + y + z + 1 = 0, 2x – y + z + 3 = 0 is :

If for some $$\alpha $$ $$ \in $$ R, the lines

L₁ : $${{x + 1} \over 2} = {{y - 2} \over { - 1}} = {{z - 1} \over 1}$$ and

L₂ : $${{x + 2} \over \alpha } = {{y + 1} \over {5 - \alpha }} = {{z + 1} \over 1}$$ are coplanar,

then the line L₂ passes through the point :

If (a, b, c) is the image of the point (1, 2, -3) in

the line $${{x + 1} \over 2} = {{y - 3} \over { - 2}} = {z \over { - 1}}$$, then a + b + c is :

The distance of the point (1, –2, 3) from

the plane x – y + z = 5 measured parallel to

the line $${x \over 2} = {y \over 3} = {z \over { - 6}}$$ is :