 Now in welcome to the session, the given question says, find the coordinates of the points where the line x plus 1 divided by 2 is equal to y plus 2 divided by 3 is equal to z plus 3 divided by 4 meets the plane x plus y plus 4z is equal to 6. Let's start with the solution. The given line as x plus 1 divided by 2 is equal to y plus 2 divided by 3 is equal to z plus 3 divided by 4. Let this be equation number 1 and the equation of plane as x plus y plus 4z is equal to 6. Let this be equation number 2. Now let us denote equation number 1 that is let us denote each of these 3 x plus 1 divided by 2 y plus 2 divided by 3 and z plus 3 divided by 4 equal to r. Let these 3 be equal to r. So this implies that x is equal to 2r minus 1, y is equal to 3r minus 2 and z is equal to 4r minus 3. Now we have to find the coordinates of the point where the given line 1 meets the plane. So these points lie on the plane that is since these point lies on 2 therefore we have in place of x here we have 2r minus 1 plus y, y is 3r minus 2 plus 4 times of z and this is equal to 6 which further implies that 2r minus 1 plus 3r minus 2 plus 16r minus 12 is equal to 6. Now we have on adding 2r plus 3r gives 5r and 5r plus 16r gives 21r minus 1 minus 2 gives minus 3 minus 3 and minus 12 gives minus 15 and on taking on the right hand side we have 15 plus 6 which gives 21. So this implies that r is equal to 21 divided by 21 which is equal to 1. Therefore x which is equal to 2r minus 1 will be equal to 2 into 1 minus 1 which gives 1, y is equal to 3 times of 1 minus 2 which is again 1 and z is equal to 4 times of 1 minus 3 which is again 1. Therefore the coordinates of x, y and z are 1, 1, 1. Hence the line meets the plane at this point. That's our answer as hence the line meets the plane having coordinates 1, 1 and 1. So this completes the session. Bye and take care.