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(A) The equation of a plane passing through the point $(x_1, y_1, z_1)$ is given by $A(x - x_1) + B(y - y_1) + C(z - z_1) = 0$.Substituting the point

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$8x + 14y + 13z + 37 = 0$

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(A) The equation of a plane passing through the point $(x_1, y_1, z_1)$ is given by $A(x - x_1) + B(y - y_1) + C(z - z_1) = 0$.Substituting the point $(2, -1, -3)$,we get $A(x - 2) + B(y + 1) + C(z + 3) = 0$.Since the plane is parallel to the lines with direction ratios $(3, 2, -4)$ and $(2, -3, 2)$,the normal vector $(A, B, C)$ must be perpendicular to both direction vectors.Thus,$3A + 2B - 4C = 0$ and $2A - 3B + 2C = 0$.Using the cross product to find the normal vector $(A, B, C)$:$A = (2)(2) - (-3)(-4) = 4 - 12 = -8$$B = -((3)(2) - (2)(-4)) = -(6 + 8) = -14$$C = (3)(-3) - (2)(2) = -9 - 4 = -13$So,the normal vector is $(-8, -14, -13)$.Substituting these into the plane equation: $-8(x - 2) - 14(y + 1) - 13(z + 3) = 0$.Multiplying by $-1$,we get $8(x - 2) + 14(y + 1) + 13(z + 3) = 0$.Expanding this: $8x - 16 + 14y + 14 + 13z + 39 = 0$.Simplifying,we get $8x + 14y + 13z + 37 = 0$.

The equation of the plane passing through the point $(2, -1, -3)$ and parallel to the lines $\frac{x - 1}{3} = \frac{y + 2}{2} = \frac{z}{-4}$ and $\frac{x}{2} = \frac{y - 1}{-3} = \frac{z - 2}{2}$ is

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