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(C) The equation of any plane passing through the intersection of the planes $P_1: x+y+z-1=0$ and $P_2: 2x+3y-z+4=0$ is given by $P_1 + \lambda P_2 =

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$y-3z+6=0$

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(C) The equation of any plane passing through the intersection of the planes $P_1: x+y+z-1=0$ and $P_2: 2x+3y-z+4=0$ is given by $P_1 + \lambda P_2 = 0$. $(x+y+z-1) + \lambda(2x+3y-z+4) = 0$ $(1+2\lambda)x + (1+3\lambda)y + (1-\lambda)z + (4\lambda-1) = 0$. Since the plane is parallel to the $x$-axis,its normal vector $\vec{n} = (1+2\lambda, 1+3\lambda, 1-\lambda)$ must be perpendicular to the $x$-axis direction vector $\vec{i} = (1, 0, 0)$. Therefore,the dot product is zero: $(1+2\lambda)(1) + (1+3\lambda)(0) + (1-\lambda)(0) = 0$. $1+2\lambda = 0 \Rightarrow \lambda = -\frac{1}{2}$. Substituting $\lambda = -\frac{1}{2}$ into the equation: $(1+2(-\frac{1}{2}))x + (1+3(-\frac{1}{2}))y + (1-(-\frac{1}{2}))z + (4(-\frac{1}{2})-1) = 0$. $0x + (1-\frac{3}{2})y + (1+\frac{1}{2})z + (-2-1) = 0$. $-\frac{1}{2}y + \frac{3}{2}z - 3 = 0$. Multiplying by $-2$,we get $y - 3z + 6 = 0$.

The equation of the plane through the intersection of the planes $x+y+z=1$ and $2x+3y-z+4=0$ and parallel to the $x$-axis is:

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