The lines $x-y+z=1,3x-y-z+1=0\text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}}x+4y-z=1,x+2y+z+1=0$are

detailed solution

Correct option is D

The direction ratios of line of intersection of two planes x−y+z=1,3x−y−z+1=0are proportional to the vector ijk1−113−11=i0−j−2+k2The direction ratios of the line L1is0,1,1A point on the line P0,−1,0, by substituting  z=0solve the simultaneous equations. The direction ratios of line of intersection of two planes x+4y−z=1,x+2y+z+1=0are proportional to the vector  ijk14−1121=i6−j2+k−2The direction ratios of the line L2 is 3,−1,−1 A point on the line is Q−3,1,0, by substituting  z=0solve the simultaneous equations.To check whether the lines lies in the same plane or not , find the value of x2−x1y2−y1z2−z1a1b1c1a2b2c2Hence, x2−x1y2−y1z2−z1a1b1c1a2b2c2= 00−21213−1−1=14, not equal to zero.Therefore, the lines are skew lines and not perpendicular.