if u've not got it, here's it is:

 

consider the foot of hill on gun's side as origin

slope of hill's surface = 30deg

so its eqn is y=xtan30

or y=x/^{[ ]}3                                                             ....(i)

 

now consider the trajectory of projectile

its eqn is

y=xtan - gx²/2u²cos²

 

now put the values of g=10, u=21,  =60 deg (the angle of projection is 60 deg)

 

the eqn becomes

y=3x-20x²/441                                                         ....(ii)

 

the point on the hill where the shell will fall will be the intersection of eqns (i) and (ii)

 

solving both, we get x=441/103 and y=441/30 m

so its distance frm the origin = (x²+y²) 

which u get as 441/15 m