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The given problem involves two parts: obtaining the perspective projection transformation for a plane and performing shearing on a square. Let's solve step by step. Perspective Projection Transformation The centre of projection is at the origin (O(0,0,0)). The projection plane passes through (P(1,5,6)) and has a normal vector (\mathbf{n} = (3,2,-1)). The general equation of a plane is: [ n_x x + n_y y + n_z z + d = 0 ] Substitute the point (P(1,5,6)) to find (d): [ 3(1) + 2(5) + (-1)(6) + d = 0 \implies 3 + 10 - 6 + d = 0 \implies 7 + d = 0 \implies d = -7 ] So, plane equation: [ 3x + 2y - z - 7 = 0 ] For _________ _______ __________ ______ ________ __________ ____ ____ ______ _______.
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