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(D) The equation of any plane passing through the intersection of planes $P_1$ and $P_2$ is given by $P_1 + \lambda P_2 = 0$.$(3x + 2y + 5z + 1) + \la

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$2x + 5y - 4z + 19 = 0$

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(D) The equation of any plane passing through the intersection of planes $P_1$ and $P_2$ is given by $P_1 + \lambda P_2 = 0$.$(3x + 2y + 5z + 1) + \lambda(x + y + z + 2) = 0$$(3 + \lambda)x + (2 + \lambda)y + (5 + \lambda)z + (1 + 2\lambda) = 0$Since this plane is parallel to the line $L$ with direction ratios $(1, 2, 3)$,the normal to the plane must be perpendicular to the line. Thus,the dot product of the normal vector $(3 + \lambda, 2 + \lambda, 5 + \lambda)$ and the direction vector of the line $(1, 2, 3)$ is zero.$1(3 + \lambda) + 2(2 + \lambda) + 3(5 + \lambda) = 0$$3 + \lambda + 4 + 2\lambda + 15 + 3\lambda = 0$$6\lambda + 22 = 0$$6\lambda = -22 \Rightarrow \lambda = -\frac{11}{3}$Substituting $\lambda = -\frac{11}{3}$ into the equation of the plane:$(3 - \frac{11}{3})x + (2 - \frac{11}{3})y + (5 - \frac{11}{3})z + (1 - \frac{22}{3}) = 0$$-\frac{2}{3}x - \frac{5}{3}y + \frac{4}{3}z - \frac{19}{3} = 0$Multiplying by $-3$,we get $2x + 5y - 4z + 19 = 0$.

Find the equation of the plane passing through the intersection of the planes $P_1$ and $P_2$ and parallel to the line $L$,where:
$P_1 : 3x + 2y + 5z + 1 = 0$
$P_2 : x + y + z + 2 = 0$
$L : \frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 3}{3}$

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Find the equation of the plane passing through the intersection of the planes $P_1$ and $P_2$ and parallel to the line $L$,where:$P_1 : 3x + 2y + 5z + 1 = 0$$P_2 : x + y + z + 2 = 0$$L : \frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 3}{3}$