Find the equation of the plane containing the lines--

r = i + j + u(i + 2j - k) and r = i + j + k(-i + j - 2k)

Find the distance of this plane from the origin and also from the point (1,1,1)

Here your line equation are not clear, so i am assuming it to be like this:
r = i + j + k + u(i + 2j - k) and r = i + j + k + t(-i + j - 2k)

We need to find the equation of the plane containing the the given to lines.
So the normal vector to the required plane can be found by the cross product of the vector equation of the two lines as:

i(-4 +1) -j(-2-1) + k (1 +2)
= -3i + 3j + 3k
So this is a normal vector passing through (1,1,1)
Hence the plane is (r-a).n
= {r - (i + j + k).}.-3i + 3j + 3k = 0
Let r = xi + yj + zk
So -3x + 3y  + 3z -3 = 0 is the required equation of the plane.
And distance  d of a point from the plane is given by d = $\frac{\left|A{x}_1+B{y}_1 +C{z}_1+D\right|}{\sqrt{A^2+B^2+C^2}}$
Here A = -3, B = 3, C = 3, D = -3
Distance from origin = $\frac{\left|-3\left(0\right) +3\left(0\right) +3\left(0\right) -3\right|}{\sqrt{3^2+3^2+3^2}}=\frac{3}{3\sqrt{3}}=\frac{1}{\sqrt{3}}$
And distance from (1,1,1) = 0 as the plane is passing through (1,1,1)