[QUESTIONPART4] Kindly provide the correct and complete solution. (I will like your solution if it is complete + Please follow the sample format below) Examples:1. Let 2 be a plane containing the point Po(4, -2, -3) and having a normal vectorN = (2,-1,2).The standard equation of the plane 2 is given by 2 (x-4)-1(y + 2) + 2(+3) = 0.So, the general equation of the plane 2 is given by 2x-y+2z-4 = 0.To sketch the graph of the plane, we use the intercepts:2-intercept: 2y-intercept:2-intercept: 2Po2. In geometry, it is known that a plane is uniquely determined by any three noncollinearpoints of the plane.Let I be a plane containing the points P(2,3,0), Q(0,5, -1) and R(1,0,3).To determine the equation of the plane, we need to obtain a normal vector-a vector thatis orthogonal to the plane. This can be obtained by solving for a vector that is orthogonalto vectors determined by points P, Q and R.We consider vectors PQ = (-2,2,-1) and PR = (-1,-3,3). A normal vector to theplane should be orthogonal to both vectors PQ and PR. This can then be obtained usingcross product.XLet N = PQ PR. Vector N is orthogonal to the plane. So, N is a normal vector to I.i jComputing, N=PQ x PR = -2 2k-1=(3,7,8).-1-33Using N as a normal vector and Po any of the three given points, a standard equationcan be obtained. The resulting general equation of plane I is 3x + 7y +82-27 = 0.Verify that points P, Q and R each satisfies the equation.Alternatively, obtain a normal vector to the plane using the cross product of vectors QRand QP. Use this normal vector to obtain a standard equation of the plane. Verify thatthis leads to the same eral equation of plane I. Determine the general equation of plane I passing through the point (-4,3,-1)and parallel to the plane 2r-y-3z +4= 0. Also, sketch the graph of this plane.

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Determine the general equation of plane I passing through the point (-4, 3,-1) and parallel to the plane 2x − y − 3z +4= 0. Also, sketch the graph of this plane.