Equation of given straight lines and ​L1​:1x−1​=−1y​=3z−1​L2​:−3x−1​=−1y​=1z−1​​having vector form respectively, are and r=(i^+k^)+v(−3i^−j^​+k^)∵(i^−j^​+3k^)⋅(−3i^−j^​+k^)=−3+1+3=1 is positive, ​r=(i^+k^)+λ(i^−j^​+3k^)∴ Angle between supporting line vectors of lines L1​ and L2​ is acute, and point of intersection of given lines L1​ and L2​ is (1,0,1).Now, vector along the acute angle bisector of vectors (i^−j^​+3k^) and (−3i^−j^​+k^) is (−i^−j^​+2k^) or (i^+j^​−2k^).L:1x−α​=y−1​=γz−γ​ is the bisector of the acute anglebetween the lines L1​ and L2​, so and ⇒​11−α​=10−1​=−21−γ​α=2,γ=−1​