 Hello and welcome to the session. In this session we will discuss a question which says that shown that the equation Ax plus Cz plus D is equal to 0 represents a plane parallel to the axis O y. Finally, the equation of the plane through the points 3 minus 2 5 and 1 minus 3 4 and parallel to O y. Now before starting the solution of this question we should know our result. And that is our perpendicular to each other even into m2 plus m1 into m2 is equal to 0. There l1, m1 and m1 are the direction cosines of one line and l2, m2 and m2 are the direction cosines of the other line. Now this result will work out as a key idea for solving out this question. And now we will start with the solution. Now in this question we have to show that the equation Ax plus Cz plus D is equal to 0 represents a plane parallel to the axis which will show that the normal to the given plane that is this plane is equal to the axis O y. If the normal to the given plane is perpendicular to the axis O y then this plane to the axis O y. Now given x plus Cz plus D is equal to 0. We know that the cosines 0 1 0 the direction cosines to the normal to the plane B. Now let us name this as equation A. Now for this plane which is given by equation A that is for the given plane 0 this condition of perpendicularity for proving that the normal to the given plane is perpendicular to the axis O y. Now let these direction cosines be l1, m1 and m1. And these direction B is equal to 0 into l2 that is 0 into A that is 1 into 0 that is 0 into 0. Now in solving this this will be equal to plus 1 into 0 plus 0 into C is equal to 0. Therefore the normal is perpendicular, condition of perpendicularity the given plane along to the axis. Now we have to find the equation of the plane through the point. Now l1 into the axis O plus Cz plus D is equal to 0. Now it was also given that the points of the plane will be now have axis 3 and that is 5. So the equation of the plane will become plus D is equal to 0. The equation of the plane will be 4C plus D is equal to 0. Now let it be equation number 1 this be equation number 2. Now solving by the method of cross multiplication minus 4 into 1 that is 5 minus 4 is equal to C over 1 into 1 that is 1 that is 3 is equal to D over l minus 5 into 1 is equal to. Now the simple to C over minus 2 is equal to D over 7 is equal to k which further and D is equal to 3. Now putting the values and D in equation number 3 we got 7k is equal to 0 which further and 2 z plus 7 is equal to 0 which is the required equation of the given equation and that's all for this session. Hope you all have enjoyed the session.