 Hello and welcome to the session. In this session we discussed the following question which says, a variable plane is at a constant distance 3p from the origin, cuts the coordinate axis at abc by the locus of the centroid of triangle abc. Let's proceed with the solution now. First of all we take net the equation of the variable plane bx upon a plus y upon b plus z upon c equal to 1. Let this be equation 1. Now we have plane 1 needs to be x axis, y axis, z axis at point a with coordinates a00, b with coordinates 0b0 and c with coordinates 00c respectively. Then we take net the coordinates of the centroid of triangle abc b alpha beta gamma. So this means we have alpha is equal to the x coordinate of point a which is a plus the x coordinate of point b which is 0, plus the x coordinate of point c which is 0 upon 3 that is equal to a upon 3. Then beta is equal to y coordinate of point a that is 0 plus the y coordinate of point b which is b, plus the y coordinate of point c which is 0 upon 3 this is equal to b upon 3. Then gamma is equal to the z coordinate of point a that is 0 plus the z coordinate of point b which is 0, plus the z coordinate of point c which is c upon 3 and this is equal to c upon 3. So this means we get a is equal to 3 alpha, b is equal to 3 beta and c is equal to 3 gamma. We are given that 3p is the distance of the plane 1 from the origin that is 000 that is we have 3p is equal to, now let's find out the distance of the plane 1 from the origin. So this would be equal to modulus of 0 upon a plus 0 upon b plus 0 upon c minus 1 and this whole upon square root of 1 upon a square plus 1 upon b square plus 1 upon c square. Further we get 3p is equal to 1 upon square root of 1 upon a square plus 1 upon b square plus 1 upon c square. So we get square root of 1 upon a square plus 1 upon b square plus 1 upon c square is equal equal to 1 upon 3 p. Now squaring both sides we get 1 upon a square plus 1 upon b square plus 1 upon c square is equal to 1 upon 9 p square. Now in place of a, b, c here we put 3 alpha 3 beta 3 gamma respectively. So this would mean that 1 upon 3 alpha square plus 1 upon 3 beta square plus 1 upon 3 gamma square is equal to 1 upon 9 p square. So we have 1 upon 9 alpha square plus 1 upon 9 beta square plus 1 upon 9 gamma square is equal to 1 upon 9 p square. So from here we have 1 upon alpha square plus 1 upon beta square plus 1 upon gamma square equal to 1 upon p square that is alpha to the power minus 2 plus beta to the power minus 2 plus gamma to the power minus 2 is equal to p to the power minus 2. Hence we say the required locus is x to the power minus 2 plus y to the power minus 2 plus z to the power minus 2 equal to p to the power minus 2. So this is our final answer. This completes the session. Hope you have understood the solution of this question.