If the lines 2x+y−k=0,x−y+1=0 and −2x−3y+8=0 are concurrent then k=

The condition that the lines a1x+b1y+c1=0,a2x+b2y+c2=0 and a3x+b3y+c3=0 are to  be concurrent is a1    b1    c1a2    b2    c2a3    b3    c3=0Hence,21−k1−11−2−38=0Expand the determinant2−8+3−18+2−k−3−2=0−10−10+5k=0k=205=4 Therefore, k=4