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(D) The equation of the family of planes passing through the line of intersection of the planes $x+y+z-1=0$ and $2x+3y+4z-5=0$ is given by $(x+y+z-1)

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$\overline{r} \cdot(\hat{i}-\hat{k})+2=0$

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(D) The equation of the family of planes passing through the line of intersection of the planes $x+y+z-1=0$ and $2x+3y+4z-5=0$ is given by $(x+y+z-1) + \lambda(2x+3y+4z-5) = 0$. Rearranging the terms,we get $(1+2\lambda)x + (1+3\lambda)y + (1+4\lambda)z - (1+5\lambda) = 0$. This plane is perpendicular to the plane $x-y+z=0$. The normal vectors of these two planes are $\vec{n_1} = (1+2\lambda)\hat{i} + (1+3\lambda)\hat{j} + (1+4\lambda)\hat{k}$ and $\vec{n_2} = \hat{i} - \hat{j} + \hat{k}$. Since the planes are perpendicular,their dot product is zero: $\vec{n_1} \cdot \vec{n_2} = 0$. $(1+2\lambda)(1) + (1+3\lambda)(-1) + (1+4\lambda)(1) = 0$. $1+2\lambda - 1 - 3\lambda + 1 + 4\lambda = 0$. $1 + 3\lambda = 0 \Rightarrow \lambda = -\frac{1}{3}$. Substituting $\lambda = -\frac{1}{3}$ into the equation of the plane: $(1 - \frac{2}{3})x + (1 - 1)y + (1 - \frac{4}{3})z - (1 - \frac{5}{3}) = 0$. $\frac{1}{3}x + 0y - \frac{1}{3}z + \frac{2}{3} = 0$. Multiplying by $3$,we get $x - z + 2 = 0$. The vector equation is $\overline{r} \cdot (\hat{i} - \hat{k}) + 2 = 0$.

The vector equation of the plane passing through the line of intersection of the planes $x+y+z=1$ and $2x+3y+4z=5$,which is perpendicular to the plane $x-y+z=0$,is

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