text: There are various ways you can proceed.\n\nr(1)=(0,1,-2) + t(1,-2,4) and r(2)=(-5,3,1) + t(1,-2,4)\n\nOne way is to find a normal to the plane first and then a point on the plane and write the equation in point normal form.\nThe thing to understand here is that by giving various values to t you obatin various point all of them on the plane.\n\nSo let us find two points on r(1) and one point on r(2)\nTwo points on r(1) put t=0,1 you have \n(0,1,-2),(1,-1,2) \nand r(2) put t=0 you have\n (-5,3,1)\nSo you have three points on a plane which are not all collinear because we chose one of them from second line.\nSo you can write the standard equation of plane passing through these three points.\nIf you need more help then note that\nIf you name these three points as r1,r2,r3 then \n(r1-r3)cross(r2-r3) = vector normal to the plane\n(5,-2,-3)cross(6,-4,1)=vector normal to the plane\nform the determinant find the normal.\nThen you have the normal take any of the points r1,r2,r3 and form the equation in point normal form.