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

Vedclass · Answer

(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 these direction vectors. Thus,$3a + 2b - 4c = 0$ and $2a - 3b + 2c = 0$. Using the cross product to find the normal vector $(a, b, c) = (3, 2, -4) 	imes (2, -3, 2) = \begin{vmatrix} i & j & k \ 3 & 2 & -4 \ 2 & -3 & 2 \end{vmatrix} = i(4 - 12) - j(6 + 8) + k(-9 - 4) = -8i - 14j - 13k$. Taking the normal vector as $(8, 14, 13)$,the equation of the plane is $8(x - 2) + 14(y + 1) + 13(z + 3) = 0$. 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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