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

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$33x+45y+50z-41=0$

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(B) The equation of the family of planes passing through the intersection of the planes $x+2y+3z-4=0$ and $2x+y-z+5=0$ is given by $(x+2y+3z-4) + \lambda(2x+y-z+5) = 0$. Rearranging the terms,we get $(1+2\lambda)x + (2+\lambda)y + (3-\lambda)z + (-4+5\lambda) = 0$ ... $(1)$. Since this plane is perpendicular to the plane $5x+3y-6z+8=0$,the dot product of their normal vectors must be zero. Thus,$(1+2\lambda)(5) + (2+\lambda)(3) + (3-\lambda)(-6) = 0$. Expanding this,we get $5 + 10\lambda + 6 + 3\lambda - 18 + 6\lambda = 0$. Simplifying,$19\lambda - 7 = 0$,which gives $\lambda = \frac{7}{19}$. Substituting $\lambda = \frac{7}{19}$ into equation $(1)$: $(1 + 2(\frac{7}{19}))x + (2 + \frac{7}{19})y + (3 - \frac{7}{19})z + (-4 + 5(\frac{7}{19})) = 0$. $(\frac{19+14}{19})x + (\frac{38+7}{19})y + (\frac{57-7}{19})z + (\frac{-76+35}{19}) = 0$. $\frac{33}{19}x + \frac{45}{19}y + \frac{50}{19}z - \frac{41}{19} = 0$. Multiplying by $19$,we get $33x + 45y + 50z - 41 = 0$.

The equation of the plane that contains the line of intersection of the planes $x+2y+3z-4=0$ and $2x+y-z+5=0$ and is perpendicular to the plane $5x+3y-6z+8=0$ is

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