Cross product of #(x_1,y_1,z_1)# and #(x_2,y_2,z_2)# is given by
#(y_1z_2-y_2z_1,z_1x_2-z_2x_1,x_1y_2-y_1x_2)#
Hence cross product of #(2k,3,k+1)# and #(k-1,k,1)# is
#((3xx1-k(k+1)),((k+1)(k-1)-1xx2k),(2kxxk-3(k-1))#
or #(3-k^2-k,k^2-1-2k,2k^2-3k+3)#
As it is #(1.-2.2)#, we have
#3-k^2-k=1# i.e. #k^2+k-2=0# i.e. #k=-2# or #k=1# and
#k^2-2k-1=-2# i.e. #k^2-2k+1=0# i.e. #k=1# and
#2k^2-3k+3=2# i.e. #2k^2-3k+1=0# i.e. #k=1/2# or #k=1#
As value of #k# should hold for all components of vector.
we have #k=1#
