 Hello and welcome to the session. In this session we discussed the following question which says find the equation of the plane passing through the points 3, 4, 1 and 0, 1, 0 and parallel to the line x plus 3 upon 2 equal to y minus 3 upon 7 equal to z minus 2 upon 5. We know that equation of the plane passing through, say a point p which coordinates x1, y1, z1 is given by A into x minus x1 plus B into y minus y1 plus C into z minus z1 equal to 0 where we have a, b and c are the direction ratios of a line perpendicular to the given plane. Then consider a line say x minus x1 over a1 is equal to y minus y1 over b1 equal to z minus z1 over c1 and a plane of equation a2x plus b2y plus c2z plus d2 equal to 0. Then the given line is parallel to the given plane if a1 a2 plus b1 b2 plus c1 c2 is equal to 0. This is the key idea that we use for this question. Let us move on to the solution. First we find out the equation of the plane passing through the point 341 is given by A into x minus 3 plus b into y minus 4 plus C into z minus 1 equal to 0 or we could write this as ax plus by plus cz equal to 3a plus 4b plus c. Now let this equation of the plane be given by 1. Now we are also given in the question that plane 1 also passes through the point 010 therefore we have A into 0 plus b into 1 plus c into 0 equal to 3a plus 4b plus c that is we put x equal to 0, y equal to 1 and z equal to 0 in the given equation 1. So we get 3a plus 4b plus c minus b equal to 0 hence we have 3a plus 3b plus c equal to 0. Let this be equation 2. Now the equation of the line given to us is x plus 3 upon 2 equal to y minus 3 upon 7 equal to z minus 2 upon 5. Let this be equation 3. Now it is given in the question that plane 1 is parallel to the line 3 therefore using the key idea we get 2a plus 7b plus 5c is equal to 0. So let this be equation 4. So now we have got two equations equation 2 and equation 4 and we need to solve both these equations to get the values for a, b and c. That is we solve equation 2 and equation 4 by cross multiplication method and we get a upon 15 minus 7 equal to b upon 2 minus 15 equal to c upon 21 minus 6 that is we have a upon 8 is equal to b upon minus 13 is equal to c upon 15 let this be equal to lambda. So we get a equal to 8 lambda b equal to minus 13 lambda and c equal to 15 lambda. So now substituting the values of a, b and c in equation 1 we get 8 lambda x plus minus 13 lambda y plus 15 lambda z is equal to 3 into 8 lambda plus 4 into minus 13 lambda plus 15 lambda. So we take lambda common from both the sides and lambda lambda gets cancelled so we will be left with 8x minus 13y plus 15z is equal to 24 minus 52 plus 15 that is we get 8x minus 13y plus 15z is equal to minus 13 or we can also write this as 8x minus 13y plus 15z plus 13 is equal to 0 this is the required equation of the plane. Thus the final answer is 8x minus 13y plus 15z plus 13 equal to 0 is the equation of the plane passing through the points 3, 4, 1 and 0, 1, 0 and parallel to the given line. So this completes the session. Hope you have understood the solution for this question.