 Hello and welcome to the session. In this session we will discuss a question which says that part A find the equation of plane which passes through the point A whose coordinates are minus 2, 3, 1 and is parallel to the lines L1 and L2 whose direction ratios are 1 minus 1, 2 and 6 minus 2, 3 respectively. And B part find the equation of the plane which is the perpendicular bisector of the segment joining the point A whose coordinates are 4, 5, 6 and B whose coordinates are 2, minus 3, minus 4. Now before starting the solution of this question we should know some results. First is the equation of the plane passing through the point x1, y1, z1 is given as A into x minus x1 the whole plus B into y minus y1 the whole plus C into z minus z1 the whole is equal to 0. Where A, B and C are the constants and secondly two lines are perpendicular to each other if A1 into A2 plus B1 into B2 plus C1 into C2 is equal to 0. Where A1, B1 and C1 are the direction ratios of one line and A2, B2 and C2 are the direction ratios of the second line. Now these results will work out as a key idea for solving out this question. And now we will start with the solution. Now in the part A we have to find the equation of the plane which is passing through the point A whose coordinates are minus 2, 3, 1 and is parallel to the line cell 1 and A2 whose direction ratios are given to us. Now using this result which is given in the key idea the equation of the plane passing through the point A whose coordinates are minus 2, 3, 1 is given as A into x minus x1 the whole that is A into x minus of minus 2 which will be x plus 2 the whole plus B into y minus y1 that is y minus 3 the whole plus C into z minus z1 that is C into z minus 1 the whole is equal to 0 which implies Ax plus 2A plus B1 minus 3B plus Cz minus C is equal to 0 which implies Ax plus B1 plus Cz plus 2A minus 3B minus C is equal to 0. Now let this be equation A and this be equation 1. Now it is also given that the plane which is given by equation number 1 is parallel 1 and L2 whose direction ratios are given as 1 minus 1, 2 and 6 minus 2, 3 respectively. Now if the plane which is given by equation number 1 is parallel to the lines L1 and L2 then that means that the normal to the plane 1 is perpendicular to the lines L1 and L2. So we can write the normal to the plane which is given by equation number 1 is perpendicular to the lines L1. Now using the condition of perpendicularity if the normal to the plane which is given by equation number 1 is perpendicular to the line L1 whose direction ratios are 1 minus 1 and 2 then 1 into A2 that is A plus B1 into B2 that is B into minus 1 that is minus B plus C1 into C2 that is plus C is equal to 0. If the normal to the plane 1 is perpendicular to the line L2 whose direction ratios are given to us then by the condition of perpendicularity A1 into A2 that is 6A plus B1 into B2 that is minus 2 that is plus 3C is equal to 0. Now let this be equation number 2 and this be equation number 3 solving 2 and 3 by the method of cross multiplication A over minus 1 into 3 that is minus 3 minus of 2 that is minus of minus 4 which will be plus 4 is equal to B over 2 that is 12 minus 3 into 1 that is 3 is equal to 0 over 1 into minus 2 that is minus 2 minus of 6 into minus 1 that is minus of minus 6 that is plus 6 is equal to now that these all equal to the constant k. Now this implies A over 1 is equal to B over 9 is equal to C over 4 is equal to k which further implies A is equal to k, B is equal to 9k and C is equal to 4k. Now this is the equation A now putting the values of A, B and C in equation A we get k into plus 2 the whole plus 9k into y minus 3 the whole plus 4k into z minus 1 the whole is equal to 0 which implies now dividing through and yk and further solving it will be x plus 2 minus 27 plus 4z is equal to 0 which further implies plus 9y plus 4z minus 29 is equal to 0 the required equation now let us start with the B part. Now in the B part we have to find the equation of the plane which is the perpendicular bisector of the segment joining A and B. Now let us play is the perpendicular bisector of the segment joining the points A and B. Now the direction which shows the line AB 4 minus 2 minus of minus 3 that is 5 plus 3 6 minus of minus 4 that is 6 plus 4 so this is equal to 2 8 and 10 or you can write 1 4 and now it is given that this plane is the perpendicular bisector of the line segment AB this implies that the plane will divide the line segment AB into two equal segments so let us take this point as the point m which is the midpoint of the line segment AB. Now our midpoint formula coordinates of the midpoint m of the line segment AB are 4 over 2 5 plus of minus 3 that is 5 minus 3 over 2 6 plus of minus 4 that is 6 minus 4 over 2 so this is equal to 4 plus 2 is 6 and 6 by 2 is 3 5 minus 3 is 2 and 2 by 2 is 1 and this will be equal to 1. Now this plane is the perpendicular bisector of the line segment AB this means that AB is the normal to the given plane and it also passes through the point m AB is the normal to the required plane which also passes through the midpoint m. Now the direction ratios of the line AB are 1 4 5 that means the direction ratios of the normal to the plane are 1 4 5 and it also passes through this point so now consider these direction ratios as ABC and this point that is the midpoint m as x1 y1 z1 so the required equation of the plane into x minus x1 the whole that is 1 into x minus 3 the whole plus b into y minus y1 the whole that is plus 4 into y minus 1 the whole plus b into z minus z1 the whole that is plus 5 into z minus 1 the whole is equal to 0 which implies x minus 3 plus 4 y minus 4 plus 5 z minus 5 is equal to 0 which further implies x plus 4 y plus 5 z minus 12 is equal to 0 which is the required equation of the plane. So this is the solution of the given equation and that's all for this session hope you all have enjoyed the session.