 So given a line, we can write the vector equation. For example, f of t is vector 3, 1, 4, plus a scalar multiple of the vector 1, negative 1, 4. We can convert this into the parametric equation by finding the functions for x of t, y of t, and z of t. So let's find the parametric equation of the line, and so we need a function giving us the coordinates x, y, z of any point on the line. Now, since the vector v1, v2, v3 gives the directions to the point v1, v2, v3 from the origin, then we can take our function and write it as a single vector. And we can read the coordinates because the coordinates of the point are the same as the components of the vector. So our x coordinate is 3 plus t, our y coordinate is 1 minus t, and our z coordinate is 4 plus 4t. And while technically x, y, and z are functions of t, we'll omit the of t as understood. With the parametric equations and a little bit of algebra, we can do things like find the point where the line between two points intersects a plane. So we have the equation of the plane, so we should find the equation of the line. So again, we can get to any point x, y, z on the line by traveling from the origin to the point 2, 1, 5 on the line along the vector 2, 1, 5. And then we can travel from 2, 1, 5 towards 5, negative 3, 8 along the vector 3, negative 4, 3. And so we can write the vector equation of the line, go to 2, 1, 5, and then any scalar multiple in the direction 3, negative 4, 3. And we can find the parametric equations, so our vector equation can be simplified too. And the components of the vector are the same as the coordinates of the point, x, 2 plus 3t, y, 1 minus 4t, and z, 5 plus 3t. Now we want to find the intersection of the line and the plane. So remember we're on the plane whenever 3x plus 5y plus 2z is equal to 30. Meanwhile, we're on the line whenever x equals 2 plus 3t, y equals 1 minus 4t, and z equals 5 plus 3t. And we're on both when the same values of x, y, and z make both equations true. So equals means replaceable, let's substitute and solve. So we have our equation for the plane, and we'll replace the expressions for x, y, and z, and solve. Now since we actually want the coordinates of the point, we know that t is equal to negative 9 fifths, so we'll substitute those in and find our coordinates x, y, and z, which give us our intersection point.