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(C) The equation of a plane passing through the intersection of two planes $P_1: x+2y-3z+2=0$ and $P_2: 6x+y+z+1=0$ is given by $P_1 + \lambda P_2 = 0

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$5x-y+4z=1$

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(C) The equation of a plane passing through the intersection of two planes $P_1: x+2y-3z+2=0$ and $P_2: 6x+y+z+1=0$ is given by $P_1 + \lambda P_2 = 0$. $(x+2y-3z+2) + \lambda(6x+y+z+1) = 0$ $(1+6\lambda)x + (2+\lambda)y + (-3+\lambda)z + (2+\lambda) = 0$ This plane is parallel to the line $\frac{x-1}{1} = \frac{y+2}{1} = \frac{z-7}{-1}$. The normal vector of the plane is $\vec{n} = (1+6\lambda, 2+\lambda, -3+\lambda)$ and the direction vector of the line is $\vec{v} = (1, 1, -1)$. Since the plane is parallel to the line,the normal vector is perpendicular to the direction vector,so $\vec{n} \cdot \vec{v} = 0$. $(1+6\lambda)(1) + (2+\lambda)(1) + (-3+\lambda)(-1) = 0$ $1 + 6\lambda + 2 + \lambda + 3 - \lambda = 0$ $6\lambda + 6 = 0 \Rightarrow \lambda = -1$. Substituting $\lambda = -1$ into the equation: $(1-6)x + (2-1)y + (-3-1)z + (2-1) = 0$ $-5x + y - 4z + 1 = 0$ $5x - y + 4z = 1$.

The equation of a plane passing through the intersection of two planes $x+2y-3z+2=0$ and $6x+y+z+1=0$ and parallel to the line $x-1=y+2=7-z$ is

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