 Hi and welcome to the session. Let us discuss the following question. Question says, find the Cartesian equation of the following plane. This is the given equation of the plane. First of all, let us understand that if this is the vector equation of the plane given to us, then Cartesian equation of the plane is Ax plus By plus Cz is equal to D, where A, B, C are the direction ratios of the normal to the plane. Now we will use this as our key idea to solve the given question. Let us now start with the solution. First of all, we will rewrite the given equation of the plane. It is R vector dot s minus 2t i plus 3 minus d j plus 2s plus d k is equal to 15. Clearly we can see this given equation of the plane is in this form. Now if we compare these two equations, we get A is equal to s minus 2t, B is equal to 3 minus d and C is equal to 2s plus d. Also, D is equal to 15. Now substituting these values of A, B, C and D in this equation, we get Cartesian equation of the given plane. So we can write Cartesian equation of the given plane is s minus 2t multiplied by x plus 3 minus d multiplied by y plus 2s plus d multiplied by z is equal to 15. So this is the required equation of the plane. This completes the session. Hope you understood the solution. Take care and have a nice day.