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(A) The equation of the family of planes passing through the intersection of the planes $P_1: 3x - y - 4z = 0$ and $P_2: x + 3y + 6 = 0$ is given by $

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$x - 2y - 2z - 3 = 0$,$2x + y - 2z + 3 = 0$

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(A) The equation of the family of planes passing through the intersection of the planes $P_1: 3x - y - 4z = 0$ and $P_2: x + 3y + 6 = 0$ is given by $P_1 + \lambda P_2 = 0$.$(3x - y - 4z) + \lambda(x + 3y + 6) = 0$$(3 + \lambda)x + (3\lambda - 1)y - 4z + 6\lambda = 0$ ... $(i)$The distance of this plane from the origin $(0, 0, 0)$ is given as $1$. The formula for the distance of a plane $Ax + By + Cz + D = 0$ from the origin is $\frac{|D|}{\sqrt{A^2 + B^2 + C^2}}$.$\frac{|6\lambda|}{\sqrt{(3 + \lambda)^2 + (3\lambda - 1)^2 + (-4)^2}} = 1$$|6\lambda| = \sqrt{(9 + 6\lambda + \lambda^2) + (9\lambda^2 - 6\lambda + 1) + 16}$$|6\lambda| = \sqrt{10\lambda^2 + 26}$Squaring both sides: $36\lambda^2 = 10\lambda^2 + 26$$26\lambda^2 = 26 \implies \lambda^2 = 1 \implies \lambda = \pm 1$.Case $1$: If $\lambda = 1$,equation $(i)$ becomes $(3+1)x + (3-1)y - 4z + 6 = 0 \implies 4x + 2y - 4z + 6 = 0 \implies 2x + y - 2z + 3 = 0$.Case $2$: If $\lambda = -1$,equation $(i)$ becomes $(3-1)x + (-3-1)y - 4z - 6 = 0 \implies 2x - 4y - 4z - 6 = 0 \implies x - 2y - 2z - 3 = 0$.Thus,the required planes are $x - 2y - 2z - 3 = 0$ and $2x + y - 2z + 3 = 0$.

The equation of the planes passing through the line of intersection of the planes $3x - y - 4z = 0$ and $x + 3y + 6 = 0$ whose distance from the origin is $1$,are

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