 Hello and welcome to the session. In this session we discussed the following question which says find the Cartesian equation of the plane through the intersection of the two given planes 2x plus 6y plus 12 equal to 0 and 3x minus y plus 4 z equal to 0, which will add a unit distance from the origin. Now given any two planes a1x plus b1y plus c1z plus d1 equal to 0 and a2x plus b2y plus c2z plus d2 equal to 0. Now the equation of a plane through the intersection of the given two planes is a1x plus b1y plus c1z plus d1 plus lambda into a2x plus b2y plus c2z plus d2 equal to 0. This is the key idea that we use in this question. Let's proceed with the solution now. We are given two planes 2x plus 6y plus 12 equal to 0. Let this be equation 1 and 3x minus y plus 4 z equal to 0. Let this be equation 2. So now equation of the plane through the intersection of the planes 1 and 2 is given as 6y plus 12 plus lambda into 3x minus y plus 4 z equal to 0. That is we have 2 plus 3 lambda into x plus 6 minus lambda into y plus 4 lambda into z plus 12 equal to 0. So this is the equation of the plane through the intersection of the given two planes 1 and 2. Let this be equation 3. Now we are given that this plane is at a unit distance the origin. Therefore the length of the perpendicular on the origin equation 3 or to the plane given by this equation is 1. Therefore we have 12 upon square root of 2 plus 3 lambda whole square. That is the coefficient of x whole square plus coefficient of y which is 6 minus lambda whole square plus coefficient of z which is 4 lambda whole square this is equal to 1. So further we have square root of 3 lambda whole square plus 6 minus lambda whole square plus 4 lambda whole square equal to 12. Now squaring both sides we will get 2 plus 3 lambda whole square lambda whole square plus square equal to 144. Solving further we get 4 plus 12 lambda plus 36 plus lambda square minus 12 lambda plus 16 lambda square equal to 144. This gives us 26 lambda square plus lambda cancel so 36 plus 4 is 40 is equal to 144. This gives us 26 lambda square equal to 144 minus 14. From here we get 26 lambda square is equal to 104 this means we have lambda square equal to 104 upon 26 and 26 4 times is 104. So we have lambda square equal to 4 which gives us lambda equal to plus minus 2. Now putting lambda equal to 2 in equation 3 we get 2 plus 3 into 2 into x plus 6 minus 2 into y plus 4 into 2 into z plus 12 is equal to 0. This gives us 8x plus 4y plus 8z plus 12 equal to 0. This gives us 2x plus y plus 2z plus 3 equal to 0. Now putting lambda equal to minus 2 in equation 3 we get plus 3 into minus 2 into x plus 6 minus minus 2 into y plus 2 into z plus 12 equal to 0. Which gives us would be minus 4x 6 plus 2 would be plus 8y and 4 into minus 2 is minus 8z plus 12 equal to 0. Which gives us x minus 2y plus 2z minus 3 equal to 0. So this is another equation. So the required equations are the plane plus 2z plus 3 equal to 0 plus 2y plus 2z minus 3 equal to 0. This is our final answer. This completes the session. Hope you have understood the solution of this question.