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(D) The equation of a plane passing through $(x_0, y_0, z_0)$ is $a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$. Given the point $(2, -1, -3)$,the equation is $

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

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(D) The equation of a plane passing through $(x_0, y_0, z_0)$ is $a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$. Given the point $(2, -1, -3)$,the equation is $a(x-2) + b(y+1) + c(z+3) = 0$. Since the plane is parallel to the lines with direction ratios $(3, 2, -4)$ and $(2, -3, 2)$,the normal vector $\vec{n} = (a, b, c)$ must be perpendicular to both direction vectors. Thus,$\vec{n} = (3, 2, -4) 	imes (2, -3, 2) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 2 & -4 \ 2 & -3 & 2 \end{vmatrix} = \hat{i}(4 - 12) - \hat{j}(6 + 8) + \hat{k}(-9 - 4) = -8\hat{i} - 14\hat{j} - 13\hat{k}$. Taking the normal vector as $(8, 14, 13)$,the equation becomes $8(x-2) + 14(y+1) + 13(z+3) = 0$. Expanding this,we get $8x - 16 + 14y + 14 + 13z + 39 = 0$,which simplifies to $8x + 14y + 13z + 37 = 0$.

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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