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  a1b1c1a2b2c2a3b3c3=0Hence,  21−k1−11−2−38=0  Expand the determinant 2−8+3−18+2−k−3−2=0−10−10+5k=0k=205=4Therefore, k=4