There is a simple crystallographic law:

h1*h2+k1*k2+l1*l2=0

In other words: you can define an arbitrary (hkl)2 with h2 and k2, and you can calulate l2  using the upper formulae.

Let us assume (111) = (hkl)1 then it follows:

h2+k2+l2=0, i.e., l2=-(h2+k2) 

This will be fulfilled by:

(11-2)

(12-3)

(13-4)

(27-9) etc...

BTW: This is correct for ALL crystal systems!

Thank you so much...Its very simple and good way

Let me practice it once.

If I have (110) plane

(110)=(hkl)1

therefore, h2 +k2 =0 which implies either h2=-k2 or k2=-h2

and l2 could be any no.

so possible combinations

1-10

-110

1-11

-111

1-12

-112

.......

For (100) plane

(100)=(hkl)1

therefore, h2=0 and k2 and l2 could be any value

000 % which is just a point

001

010

011

00-1

0-10

0-1-1

002

020

022

....

Am I right?

yes you are right

 Do have you information how to find direction[uvw] normal to the plane(hkl) Mr.Nolze.

can I use previous way as you mentioned above ?

Although there are many combinations of perpendicular planes for 110 planes, What should be other 2 planes perpendicular to it and simultaneously perpendicular to each other.

In simple words, If Y axis is considered as 110, what could be the possible X and Z axis?

Regarding [uvw]: This depends on the crystal system. For a triclionic crystal commonly no low-indexed [uvw] is perpendicular to an (hkl)... For a cubic crystal (hkl) and [uvw] with h=u, k=v and l=w are ALL are parallel to each other but have different length.

Of course, one can calculate also for all non-cubic systems an approximate (or exact) [uvw] || (hkl) (here the normal vector is always meant, i.e., the respective plane is perpendicular). This is commonly done by the the metric tensor G. Equations you can find in several books about crystallography or materials science.

@ Anuj Sharma: please use () and [] and not only the hkl since this nomenclature is reserved for Laue indexing and indicates the interferences of (hkl). in contrast, they can have any number whereas h,k,l in (hkl) have to prime to each other.

To your question:  (hkl)1 x (hkl)2 = [uvw]. i.e., if you take the cross product of two vectors in reciprocal space, the result is always the intersection of both planes which is a lattice vectors! Vice versa [uvw]1 x [uvw]2 = (hkl), i.e., the cross product of two lattice vectors gives a reciprocal vectors which is the normal to the plane defined by the two lattice vectors. This is valid for any crystal system. For cubic crystals it means that lattice as well as reciprocal lattice vectors with the same indices are parallel (as written above).  Regarding your question:

if you define X and Y by two lattice plane normals, their intersection vector (||Z) is ALWAYS a lattice vector and no vector in reciprocal space. Only for cubic phases this doesn't matter, and for some lower symmetries it is also valid for some selected zones, mostly [001].

If you are not working with such zones Z will be described by an exact lattive vector, but there is perhaps no low-indexed vertical plane...(sie also above the comment with the metric tensor).