<u>Solution-</u>
First of all we need co-ordinates of three points on the plane, to get the equation of that plane. We already have co-ordinate of one point on the plane, and we can get the co-ordinates of other two point from the given line equation. After getting all the 3 required co-ordinates, we can calculate two vectors on the plane and then taking their cross product will give us a normal to the desired plane.
Then we using the general equation for a plane with normal vector ,
Putting t = 0, we get a point (0, 3, 4) which is also on the plane.
Putting t = 1, we get another point (5, 4, 3) which is also on the plane.
Let a be the vector from (0, 3, 4) to (3, 4, 5)
⇒ a = < 3-0, 4-3, 5-4 > = < 3, 1 , 1 >
Let b be the vector from (5, 4, 3) to (3, 4, 5)
⇒ b = < 3-5, 4-4, 5-3 > = < -2, 0, 2 >
Taking their cross product to get the normal,
= < 2, -8, 2>
∴ Then the equation for the plane passing through (3, 4, 5) and having normal vector < 2, -8, 2 > is,