 Hello and welcome to the session, let us solve the following question which says find the vector equation of the line passing through the point with coordinates 1, 2 and 3 and parallel to the planes vector r dot i cap minus j cap plus 2 k cap is equal to 5 and vector r dot 3 i cap plus j cap plus k cap is equal to 6. Let us now start with the solution and equation of a line passing through the point whose coordinates are 1, 2 and 3 is given by i cap plus 2 j cap plus 3 k cap and the given planes vector r dot i cap minus j cap plus 2 k cap is equal to 5 and vector r dot 3 i cap plus j cap plus k cap is equal to 6. Let us denote the vector i cap minus j cap plus 2 k cap by vector b and 3 i cap plus j cap plus k cap by vector c. Now we have to find the vector equation of a line which passes through this point and is parallel to the two given planes. So this implies the line is vector b and vector c. So this implies that the line is normal to their cross product which is given by vector b cross vector c. So let us find out cross product to vector b and c which is equal to i cap j cap k cap and in this row we will write the scalar components of vector b in the direction of x, y and z which are 1 minus 1 and 2 and in this row we will write the scalar components of vector c, vector 3 1 and 1 and this is further equal to i cap and minus 1 minus 2 minus j cap 1 minus 6 plus j cap 1 plus 3. This is equal to minus 3 i cap plus 5 j cap plus 4 k cap and thus in other words we can say that the vector equation of the required line which passes through the point with condition coordinates 1, 2 and 3 normal to the vector minus 3 i cap plus 5 j cap plus 4 k cap is given by vector r which is equal to i cap plus 2 j cap plus 3 k cap plus lambda times minus 3 i cap plus 5 j cap plus 4 k cap. So this is the required equation in vector form. So this completes the session by antique care.