 Hi and welcome to the session. Let us discuss the following question. Question says, find the coordinates of the point where the line through 3-4-5 and 2-3-1 crosses the plane 2x plus y plus z is equal to 7. Let us now start with the solution. Now we know the line passes through 3-4-5 and 2-3-1. Let us name these points as a and b. So we can write line passes through point a having coordinates 3-4-5 and point b having coordinates 2-3-1. Now vector equation of the line through the points a and b is r vector is equal to 3i minus 4j minus 5k plus lambda multiplied by 2 minus 3i plus minus 3 plus 4j plus 1 plus 5k We know vector equation of a line that passes through 2 points whose position vectors are a vector and v vector is r vector is equal to a vector plus lambda multiplied by v vector minus a vector. Here we have used this equation. Clearly we can see this is the position vector of point a and this is the difference of position vector of point b and position vector of point a. Now simplifying further we get r vector is equal to 3i minus 4j minus 5k plus lambda multiplied by minus i plus j plus 6k. Now let us assume that p is the point where a-b crosses given plane. So we can write let p be the point where line a-b crosses given plane. We know given plane is 2x plus y plus z is equal to 7. Now position vector of point p must be of the form xi plus yj plus zk. Now let us name this equation as equation 1. Now we know points a, b and p are collinear. So this point must satisfy or we can say point p must satisfy equation 1. So we can write this point must satisfy equation 1. Now replacing vector r by xi plus yj plus zk. In this equation we get xi plus yj plus zk is equal to 3i minus 4j minus 5k plus lambda multiplied by minus i plus j plus 6k. Now equating the like coefficients of unit vector i, unit vector j and unit vector k in this equation we get x is equal to 3 minus lambda, y is equal to minus 4 plus lambda, z is equal to minus 5 plus 6 lambda. Now let us name this equation as equation 2, this equation as equation 3 and this equation as equation 4. Now we know these are the coordinates of point p and point p crosses the given plane. Given plane is 2x plus y plus z is equal to 7. Since point p crosses this plane so these values of x, y and z must satisfy this equation. Now let us name this equation as equation 5. Now substituting 2, 3 and 4 in 5 we get 2 multiplied by 3 minus lambda plus minus 4 plus lambda plus minus 5 plus 6 lambda is equal to 7. Now this further implies 6 minus 2 lambda minus 4 plus lambda minus 5 plus 6 lambda is equal to 7. Now solving like terms we get 5 lambda is equal to 10 dividing both the sides of this expression by 5 we get lambda is equal to 2. Now we will substitute this value of lambda in equation 2. So we can write putting lambda is equal to 2 in equation 2 we get x is equal to 1. Similarly putting lambda is equal to 2 in equation 3 we get y is equal to minus 2. Similarly substituting lambda is equal to 2 in equation 4 we get z is equal to 7. So we get the coordinates of point p are 1 minus 2, 7. So this is our required answer. This completes the session. Hope you understood the solution. Take care and have a nice day.