Final answer:
To find the parametric equations for the line through point P(2, 1, -4) that is parallel to the plane x+2y-3z=5 and perpendicular to the line x=1+s, y=-3-s, z=5+2s, we need to find the direction vector of the line and a point on the line.
Step-by-step explanation:
To find the parametric equations for the line through point P(2, 1, -4) that is parallel to the plane x+2y-3z=5 and perpendicular to the line x=1+s, y=-3-s, z=5+2s, we need to find the direction vector of the line and a point on the line.
The direction vector of the line can be found by finding the cross product of the normal vector of the plane and the direction vector of the given line. The normal vector of the plane is n = 1i + 2j - 3k and the direction vector of the given line is v = 1i - j + 2k.
Taking the cross product of n and v, we get d = -7i - 5j - 3k. So, the parametric equations for the line are x = 2 - 7t, y = 1 - 5t, and z = -4 - 3t, where t is the parameter.