vec{r} cdot(hat{i}-hat{j}+hat{k})=5 and vec{r | vedclass

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(D) The equations of the given planes are $x-y+z=5$ and $2x+y-z=3$. The equation of any plane passing through the line of intersection of these two pl

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$-4$

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(D) The equations of the given planes are $x-y+z=5$ and $2x+y-z=3$. The equation of any plane passing through the line of intersection of these two planes is given by $(x-y+z-5) + \lambda(2x+y-z-3) = 0$. Since this plane passes through the point $(0, 1, 2)$, we substitute $x=0, y=1, z=2$ into the equation: $(0-1+2-5) + \lambda(2(0)+1-2-3) = 0$. $-4 + \lambda(-4) = 0 \Rightarrow -4\lambda = 4 \Rightarrow \lambda = -1$. Substituting $\lambda = -1$ into the equation of the plane: $(x-y+z-5) - 1(2x+y-z-3) = 0$. $x-y+z-5-2x-y+z+3 = 0$. $-x-2y+2z-2 = 0 \Rightarrow -x-2y+2z = 2$. Comparing this with $\vec{r} \cdot(a\hat{i}+b\hat{j}+c\hat{k}) = m$, we get $a=-1, b=-2, c=2$. Therefore, $\frac{bc}{a^2} = \frac{(-2)(2)}{(-1)^2} = \frac{-4}{1} = -4$.

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