If the planes 2x + 3y + 4z + 7 = 0 and 4x + k | vedclass

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(D) Since the given planes $2x + 3y + 4z + 7 = 0$ and $4x + ky + 8z + 1 = 0$ are parallel,their normal vectors are proportional. Therefore,$\frac{2}{4

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$5x + 6y + 7z = 108$

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(D) Since the given planes $2x + 3y + 4z + 7 = 0$ and $4x + ky + 8z + 1 = 0$ are parallel,their normal vectors are proportional. Therefore,$\frac{2}{4} = \frac{3}{k} = \frac{4}{8}$. From $\frac{2}{4} = \frac{3}{k}$,we get $k = 6$. Now,we need to find the equation of the plane passing through the point $(k, k, k) = (6, 6, 6)$ with direction ratios of the normal as $(k-1, k, k+1) = (5, 6, 7)$. The equation of a plane passing through $(x_1, y_1, z_1)$ with normal $(a, b, c)$ is $a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$. Substituting the values,we get $5(x - 6) + 6(y - 6) + 7(z - 6) = 0$. $5x - 30 + 6y - 36 + 7z - 42 = 0$. $5x + 6y + 7z = 108$.

If the planes $2x + 3y + 4z + 7 = 0$ and $4x + ky + 8z + 1 = 0$ are parallel,then the equation of the plane passing through the point $(k, k, k)$ and having the direction ratios of its normal as $(k-1, k, k+1)$ is

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