The lineintersectsthexy-planeat (2, 7, 0), the yz-plane at (0, 9, -6), and the xz-plane at (9, 0, 21).To findparametricequations for the line through (4, 5, 6) that is perpendicular to the plane x - y + 3z = 2, we can use the normal vector of the plane to determine thedirectionof the line.The normalvectorof the plane x - y + 3z = 2 is (1, -1, 3). Since the line is perpendicular to the plane, the direction vector of the line will be parallel to the normal vector of the plane.Let's use theparametert to represent the line. We can set up the parametricequationsas follows:x(t) = 4 + ty(t) = 5 - tz(t) = 6 + 3tThe directionvector(1, -1, 3) is multiplied by the parameter t to determine the change in x, y, and z coordinates from the initial point (4, 5, 6).(a) xy-plane (x, y, z) = (x(t), y(t), z(t)) = (4 + t × 1, 5 - t × 1, 6 + t × 3)Setting z(t) = 0, we get:6 + t × 3 = 0t = -2Substituting t = -2 into the parametric equations, we get:(x, y, z) = (4 - 2, 5 + 2, 6 - 2 × 3)= (2, 7, 0)(b) yz-plane (x, y, z) = (x(t), y(t), z(t))= (4 + t×1, 5 - t × 1, 6 + t × 3)Setting x(t) = 0, we get:4 + t × 1 = 0t = -4Substituting t = -4 into theparametricequations, we get:(x, y, z) = (4 - 4, 5 + 4, 6 - 4 × 3)= (0, 9, -6)(c) xz-plane (x, y, z) = (x(t), y(t), z(t)) = (4 + t × 1, 5 - t × 1, 6 + t × 3)Setting y(t) = 0, we get:5 - t × 1 = 0t = 5Substituting t = 5 into the parametric equations, we get:(x, y, z) = (4 + 5, 5 - 5, 6 + 5 × 3)= (9, 0, 21)To learn more about theparametric equations;brainly.com/question/29187193#SPJ12...

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