 Hello and welcome to the session. In this session, we discuss the following question that says write the vector equation of the following line x minus 5 upon 3 is equal to y plus 4 upon 7 is equal to 6 minus z upon 2. So, we are given this equation that is the Cartesian equation of the line and we are supposed to find the vector equation of this line. The Cartesian equation as the line is given by x minus x1 upon a is equal to y minus y1 upon b is equal to z minus z1 upon c. This shows that the line passes through the point a with coordinates x1, y1, z1 and this adc are the direction ratios of the line. Then the vector equation of the line is given as vector r is equal to vector r1 plus lambda into vector r2 where this lambda is some real number and vector r1 would be given by x1 i cap plus y1 j cap plus z1 k cap. So, y1 z1 are the coordinates of the point through the line passes and vector r2 would be equal to ai cap plus bj cap plus ck cap where this adc are the direction ratios of the given line. The given line parallel to this is the key idea that we use for this question. Let us now proceed with the solution. We are given the Cartesian form of the equation of the line as x minus 5 upon 3 equal to y plus 4 upon 7 is equal to 6 minus z upon 2. Let this be equation 1 or we can also write this as x minus 5 upon 3 is equal to y minus of minus 4 upon 7 is equal to z minus 6 upon minus 2. So, this means that the given line if it coordinates 5 minus 4 6 direction ratios. The position vector of the point a be given as r1 and this would be equal to 5i cap minus 4j cap plus 6k cap and also given by 3i cap plus 7j cap 2k cap. The vector r is equal to vector r1 plus lambda into vector r2 where this lambda is some real number. So, from this we can say vector r is equal to 4j cap plus 6k cap plus lambda into 3i cap plus 7j cap minus 2k cap be whole. So, this is the vector equation of the given line. This is our final answer. This completes the session. Hope you have understood the solution of this question.