Equation of L1 is
x−00=y−11=z−1−1
Similarly, equation of L2 is x−11=y−00=z+11
So, equation of any plane through line L1 is ax+b(y−1)+c(z−1)=0 ⋯(1)
where a×0+b×1+c×(−1)=0 ⋯(2)
Also, the plane (1) is parallel to the line L2.
So a×1+b×0+c×1=0 ⋯(3)
From (2) and (3), we get
a1=b−1=c−1
So, the required plane P, is
1(x−0)−1(y−1)−1(z−1)=0
⇒x−y−z+2=0
Distance from the origin to the plane x−y−z+2=0 is
d=∣∣∣2√1+1+1∣∣∣=2√3
Hence, 3d2=4