 Hello and welcome to the session. Let us discuss the following question. Question says, find the vector equation of the line passing through 1, 2, 3 and perpendicular to the plane r vector dot i plus 2j minus 5k plus 9 is equal to 0. First of all let us understand that vector equation of the line that passes through a given point whose position vector is a vector and parallel to a given vector b is r vector is equal to a vector plus lambda multiplied by p vector. This is the key idea to solve the given question. Let us now start with the solution. Now we are given that line passes through point 1, 2, 3 and also line is perpendicular to the plane r vector dot i plus 2j minus 5k plus 9 is equal to 0. Now let us assume that a vector represents the position vector of this point. So we can write a vector is equal to i plus 2j plus 3k. Clearly we can see these are the components of vector a and vector a represents the position vector of this point. Now clearly we can see given equation of the plane is in the form r vector dot n vector is equal to d here n vector is normal to the plane. Now comparing these two equations we get n vector is equal to i plus 2j minus 5k Now we know this normal vector is perpendicular to the plane and plane is perpendicular to the line. This implies this normal vector is parallel to the line. So we can write normal vector is perpendicular to the plane and plane is perpendicular to the line. Now this implies normal vector i plus 2j minus 5k is parallel to the line. Now we get direction ratios of the line are same as the direction ratios of this normal vector. Now from key idea we know vector equation of a line is r vector is equal to a vector plus lambda multiplied by b vector where a vector is the position vector of any point on the line and b vector is parallel to the line. Now here this is the position vector of the point on the line and this is the vector parallel to the line. Now we get required equation of the line is r vector is equal to i plus 2j plus 3k plus lambda multiplied by i plus 2j minus 5k. Here these are the components of vector a. We know vector a is the position vector of a point on the line and these are the components of vector parallel to the line. So using this equation we get the required equation of the line. So this is our required answer. This completes the session. Hope you understood the solution. Take care and have a nice day.