 Good morning friends. I am Purva and today we will work out the following question find the vector equation of a plane Which is at a distance of 7 units from the origin and normal to the vector 3i cap plus 5j cap minus 6k cap Now the vector equation of a plane Which is at a distance of d units from origin normal to vector n is given by Vector r dot n cap is equal to d where n cap is equal to vector n upon magnitude of vector n. So this is the key idea behind our question Let us begin with the solution now Now we are given that The plane is normal to the vector 3i cap plus 5j cap minus 6k cap. So let Vector n is equal to 3i cap plus 5j cap minus 6k cap then we have n cap which is equal to vector n upon magnitude of vector n is Equal to now vector n is equal to 3i cap plus 5j cap minus 6k cap Upon magnitude of vector n is given by under root of 3 square plus 5 square plus minus 6 whole square and this is equal to 3i cap plus 5j cap minus 6k cap upon under root of 3 square is 9 plus 5 square is 25 plus minus 6 whole square is equal to 36 And this is equal to 3i cap plus 5j cap minus 6k cap upon under root of 70. Now we have to find the vector equation of a plane which is at a distance of seven units from the origin So we are given d is equal to 7 Thus The required vector equation of the plane is Vector r dot n cap is equal to d that is Vector r dot now n cap is equal to 3i cap plus 5j cap minus 6k cap upon under root 70 Is equal to d which is equal to 7 Thus we have got the required vector equation of the plane as vector r dot 3i cap plus 5j cap minus 6k cap upon under root 70 is Equal to 7. This is our answer. Hope you have understood the solution. Bye and take care