 Hello friends, let's work out the following problem. It says find the equation of the plane passing through the points 1, 2, 3 and 0 minus 1, 0 and parallel to the line x minus 1 by 2 is equal to y minus 2 by 3 is equal to z upon minus 3. So let's now move on to the solution. We are given that the plane for which we have to find the equation passes through the points 1, 2, 3 and 0 minus 1, 0. So the equation of the plane passing through 1, 2, 3 is given by a into x minus 1 plus b into y minus 2 plus c into z minus 3 is equal to 0. As we know that the equation of the plane passing through the point a 1, a 2, a 3 is given by a into x minus a 1 plus b into x minus a 2 plus c into x minus a 3 is equal to 0. Now this plane also passes through the points 0 minus 1, 0. That means this point will satisfy the equation of the plane. So we have a into 0 minus 1 plus b into minus 1 minus 2 plus c into 0 minus 3 is equal to 0. From this we have minus a minus 3 b minus 3 c is equal to 0. Taking minus common we have a plus 3 b plus 3 c is equal to 0. Let's name this as 2. Now we are given that 1 is parallel to the line x minus 1 by 2 is equal to y minus 2 by 3 is equal to z upon minus 3. Now since the plane is parallel to this line, the normal to the plane is perpendicular to the line therefore, so normal to the plane is perpendicular to the line. Therefore, a into 2 plus b into 3 plus c into minus 3 is equal to 0. Now this equation becomes 2 a plus 3 b minus 3 c is equal to 0. Let's name this as 4. Now we'll solve equation 2 and 4 for a, b, c using method of cross multiplication. If we have equation of the form a1x plus a2y plus a3z is equal to 0 and b1x plus b2y plus b3z is equal to 0 then we have x upon a2 into b3 minus b2 into a3 is equal to y upon b1 into a3 minus a1 into b3 is equal to z upon a1 into b2 minus b1 into a2. So solve 2 and 4 using cross multiplication. So we have a upon minus 3 into 3 minus 3 into 3. Similarly, b upon 2 into 3 minus minus 3 into 3 is equal to c upon 1 into 3 minus 2 into 3. So we have a upon minus 18 is equal to b upon 15 is equal to c upon 3 minus 6 that is minus 3. Let this be equal to lambda. So we have a is equal to minus 18 lambda b is equal to 15 lambda c is equal to minus 3 lambda. Now substitute values of a, b and c in 1. Now this was equation 1 so we have a into x minus 1 that is minus 18 lambda into x minus 1 plus b into y minus 2 that is 15 lambda into y minus 2 plus c into z minus 3 that is minus 3 lambda into z minus 3 is equal to 0. Taking 3 lambda common we have we take minus 3 lambda common. So we have 6 into x minus 1 plus 5 into since we are taking minus 3 lambda common sign will change it becomes minus 5 into y minus 2 plus z minus 3 is equal to 0. Now since 6 into x is 6x minus 6 minus 5y plus 10 plus z minus 3 is equal to 0. So this implies 6x minus 5y plus z plus 1 is equal to 0 and this is the required equation of the plane. So this completes the question and the session. Bye for now. Take care. Have a good day.