 Hi and welcome to the session. I am Shashi and I am going to help you with the following question. Question says, find the coordinates of the point where the line through 5, 1, 6 and 3, 4, 1 crosses the YZ plane. First of all let us understand that vector equation of a line that passes through two points whose position vectors are A vector and B vector is R vector is equal to A vector plus lambda multiplied by B vector minus A vector. This is the key idea to solve the given question. Let us now start with the solution. Now first of all let us assume that the point A has coordinates 5, 1, 6 and point B has coordinates 3, 4, 1. So we can write line passes through A having coordinates 5, 1, 6 and point B having coordinates 3, 4, 1. Now position vector of point A is vector A which is equal to 5I plus J plus 6K. Now position vector of point B is represented by vector B which is equal to 3I plus 4J plus K. Now from key idea we know vector equation of a line passing through two points is R vector is equal to A vector plus lambda multiplied by B vector minus A vector where A vector and B vector are position vectors of given two points. Now we know these are the components of A vector and these are the components of B vector. So we will substitute components of A vector and B vector in this equation. Now we can write R vector is equal to 5I plus J plus 6K plus lambda multiplied by B vector that is 3I plus 4J plus K minus 5K. Now this further implies R vector is equal to 5I plus J plus 6K plus lambda multiplied by minus 2I plus 3J minus 5K. Subtracting corresponding components of these two vectors we get this vector. Now let P be the point where line AB crosses YZ plane. So we can write let P be the point where line AB crosses YZ plane. Then position vector of the point P is of the form YJ plus ZK. Now let us name this equation as equation 1. Now we know points AB and P lie on the same line or we can say they are collinear. So this point must satisfy equation 1. Now replacing vector R by YJ plus ZK in this equation we get YJ plus ZK is equal to 5I plus J plus 6K plus lambda multiplied by minus 2I plus 3J minus 5K. Now equating the coefficients of unit vector I, unit vector J and unit vector K in this equation we get 5 minus 2 lambda is equal to 0, Y is equal to 1 plus 3 lambda, Z is equal to 6 minus 5 lambda. Now let us name this equation as 2, this equation as 3 and this equation as 4. Now from equation 2 we get lambda is equal to 5 upon 2. Now substituting this value of lambda in equation 3 we get Y is equal to 17 upon 2. Similarly substituting lambda is equal to 5 upon 2 in equation 4 we get Z is equal to minus 13 upon 2. Now we get the coordinates of point P are 0, 17 upon 2 minus 13 upon 2. So these are our required coordinates of the point. This is our required answer. This completes the session. Hope you understood the solution. Take care and have a nice day.