 Hi and welcome to the session, I am Shashi and I am going to help you with the following question. Question says if the points 11p and minus 301 be equidistant from the plane r vector dot 3i plus 4j minus 12k plus 13 is equal to 0 then find the value of p. Let us understand that the particular distance of a point x1, y1, z1 to the plane r vector dot n vector is equal to d is given by modulus of a vector dot n vector minus v upon modulus of n vector where vector a is the position vector of this point. Now this is our k idea to solve the given question. Let us now start with the solution. Now clearly we can see perpendicular distance of point 11p to the plane r vector dot 3i plus 4j minus 12k is equal to minus 13 is given by modulus of i plus j plus pk dot 3i plus 4j minus 12k minus minus 13 upon magnitude of n vector that is square root of 3 square plus 4 square plus minus 12 square. Clearly we can see this equation of the plane is in this form now comparing these two equations we get n vector is equal to 3i plus 4j minus 12k d is equal to minus 13 and position vector of this point is i plus j plus pk which is equal to vector a. Now substituting corresponding values of vector a vector n and d in this formula we get this expression now solving it further we get modulus of 1 multiplied by 3 plus 1 multiplied by 4 plus p multiplied by minus 12 plus 13 upon square root of 169 we know to find the dot product of these two vectors we will multiply corresponding components of these two vectors. Now this is further equal to modulus of 20 minus 12p upon square root of 169 now we know square root of 169 is equal to 13 so this expression is further equal to modulus of 20 minus 12p upon square root of 169 now let us name this expression as 1 now we will find out perpendicular distance of point minus 301 from the given plane so we can write perpendicular distance of minus 301 to plane r vector dot 3i plus 4j minus 12k is equal to minus 13 is given by modulus of minus 3i plus 0j plus k dot 3i plus 4j minus 12k minus minus 13 upon square root of 3 square plus 4 square plus minus 12 square again here we have used key idea we know this is the position vector of this given point this is vector n this is the value of d and this is the magnitude of vector n now solving this expression further we get modulus of minus 3 multiplied by 3 plus 0 multiplied by 4 plus 1 multiplied by minus 12 plus 13 upon square root of 169 we know we can find the dot product of these two vectors by multiplying their corresponding components now simplifying this expression further we get modulus of minus 8 upon 13 let us name this expression as 2 now this is given in the question that these two points are equidistant from this plane so this implies expression 1 must be equal to expression 2 now expression 1 is equal to modulus of 20 minus 12p upon 13 and expression 2 is equal to modulus of minus 8 upon 13 now multiplying both the sides of this equation by 13 we get modulus of 20 minus 12p is equal to modulus of minus 8 now this further implies 20 minus 12p is equal to 8 or 20 minus 12p is equal to minus 8 now solving this equation further we get value of p is equal to minus 12 upon minus 12 which is further equal to 1 or solving this equation we get value of p is equal to 28 upon 12 which is further equal to 7 upon 3 so required value of p is equal to 1 or 7 upon 3 so this is our required answer this completes the session hope you understood the solution take care and have a nice day