hello, i got the workings already, however i am unsure as to why each step is done, is it possible for you to explain them to me step by step with explaination and wiht drawn illustration. thanks equation is x-2 y+3z=9 (Check that the other two points also satisfy thisequation.)(b) A vector parallel to the straight line is i-2j+3k. Thus, a parametricrepresentation of the straight line isX= -2+ty= 5– 2tZ= -2+3t(c) To find the required point of intersection, substitute the above parametricequations(-2+ 1) – 2(5– 2t) + 3(-2+3t) = 9. We obtain t= 27/14. Thus, the co-ordinates ofthe point of intersection are given byforthelineintothe equation oftheplane,thatis,x= -2+ (27/14) = -1/14y= 5–2(27/14) = 8/7z=-2+3(27/14) = 53/14 2. (a) With reference to a Cartesian coordinate system 0xyz, find the plane that containsthe points (2, 1, 3), (3, 0, 2) and (-1, 1, 4) (Hint. One way of doing this is to find twovectors that lie on the plane and use them to obtain a vector perpendicular to theplane.)(b) A straight line perpendicular to the plane in part (a) passes through the point (-2,5, -2) Give a parametric representation for the straight line.(c) Find the point where the plane in part (a) and the straight line in part (b) intersect.(Solution)(a) The two vectors on the plane are:(3– 2) i+ (0–1) j+ (2– 3)k = i-j-k(3+1) i+ (0–1) j+ (2 – 4)k = 4i– j-2k%3DA cross product of the two vectors gives a normal vector to the plane. We have:(i-j-k)×(4i- j- 2k) = i– 2j+3k)Thus, the equation of the plane may be written asX- 2 y+3z= d.Use one point to work out d, e.g. use (2, 1, 3), d = 2-2+9= 9. Thus, the required

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Position vector of points, A→2,1,3 ; B→3,0,2 ; C→-1,1,4The position vectors are relative to…