Dear Ashish the solution follows:I would illustrate the solution in the way that all your concepts sould get cleared up and u are able to solve the similar probems.
Prequisites: Resolution of forces, application of 3 equations of motion with proper sign usage.
Motion analysis: the particle will strike the plane at right angle as given in the question. This is the keyword which would lead us to the solution, since at that time the component of veloity perpendicular to the incline will remain while the one parallel to the incline will become zero.
Solving Method: 1. Resolve the intitial velocity into two components, (a) parallel to plane of incline, (b) perpendicular to plane of incline.
2 Make equation of motion for two dimensions.(a) II to incline plane (b) perpendicular to incl plane
3. Apply following conditions in above written equations:
a. velocity component parallel to incline zero.
b. displacent of projectile in the plane of incline zero
4. Solve and u will get desired ans
Solution:
let, angle of projection be 
velocity components: along incline plane
acceleration component; along incline :
velocity components: perpendicular to incline plane
acceleration component; perpendicular to incline :
eqn:
1. Along incline
Ux =

2. perpendicular to the incline


now putting Ux= 0 at time of strking the plane, we get

putting y=0 in above eqn we get

equating the two t , we get

put the value of obtained t and solve for x in terms of theta and beta. then use the above relation between theta and beta. no need to take any inverse trignometric opertation. terms themselves will become simples as u proceed.
i am only writing final expression,
the height asked from horizontal will be
solving you will get the ans, note u have to first convert all trignomentric ratios in terms of sin in order to see the ans in the form written in the question.
similarly u can obtain t. if still any dobut perssits, dont hesitate to ask at all.
Vikram