From 2-D to 3-D

The problem which looks complicated can be solved easily by a specific approach which I found after solving Let`s play carrom.

Let`s assume that there are no walls and the ball is following a straight path instead of being reflected.

Also assume that there are infinite rooms,coordinates of whose corners are integers and they too have no walls!.

Now,when the ball strike corner of imaginary room then it will also strike corner of real room.

Here is an example to show my point.

So,the equation of line along which ball will travel will be given by joining

$\left( \frac { 1 }{ 4 } ,0,\frac { 1 }{ 2 } \right)$ and $\left( \frac { 5 }{ 13 } ,\frac { 3 }{ 13 } ,1 \right)$

$\frac { x-\frac { 1 }{ 4 } }{ 7 } =\frac { y }{ 12 } =\frac { z-\frac { 1 }{ 2 } }{ 26 } =r(say)$

Now, only thing left to do is to find smallest integer coordinates which will be satisfied by line.

We have x=7r+ $\frac { 1 }{ 4 }$ , y=12r , z=26r+ $\frac { 1 }{ 2}$

By finding smallest integer solution which satisfy above condition we`ll have r= $\frac { 1 }{ 4 }$ (It is easy to find try yourself)

$\therefore$ Coordinates are (2,3,7) and number of collisions will be equal to number of planes crossed =

$\boxed{2-1+3-1+7-1 =9}$ (since we do not have to count the corner in collision)

For those who are not convinced by above solution ,coordinates of points of collision are $\left( \frac { 5 }{ 13 } ,\frac { 3 }{ 13 } ,1 \right) ,\left( \frac { 17 }{ 26 } ,\frac { 9 }{ 13 } ,0 \right) ,\left( \frac { 5 }{ 6 } ,1,\frac { 2 }{ 3 } \right) ,\left( \frac { 12 }{ 13 } ,\frac { 11 }{ 13 } ,1 \right) ,\left( 1,\frac { 5 }{ 7 } ,\frac { 5 }{ 7 } \right)$ , $\left( \frac { 21 }{ 26 } ,\frac { 5 }{ 13 } ,0 \right) ,\left( \frac { 7 }{ 12 } ,0,\frac { 5 }{ 6 } \right) ,\left( \frac { 7 }{ 13 } ,\frac { 1 }{ 13 } ,1 \right) ,\left( \frac { 7 }{ 26 } ,\frac { 7 }{ 13 } ,0 \right) ,\left( 0,1,1 \right)$