If n=2 hat{i}-3 hat{j}+4 hat{k},m=hat{i}-hat{ | vedclass

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(D) The equation of a plane passing through the line of intersection of planes $r \cdot n=1$ and $r \cdot m=-4$ is given by $r \cdot (n + \lambda m) =

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$5 x-2 y-12 z+47=0$

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(D) The equation of a plane passing through the line of intersection of planes $r \cdot n=1$ and $r \cdot m=-4$ is given by $r \cdot (n + \lambda m) = 1 - 4\lambda$,where $\lambda$ is a scalar. Substituting the given vectors: $r \cdot ((2 \hat{i} - 3 \hat{j} + 4 \hat{k}) + \lambda(\hat{i} - \hat{j})) = 1 - 4\lambda$. This plane is perpendicular to the plane $r \cdot l = -8$,where $l = 2 \hat{i} - \hat{j} + \hat{k}$. Thus,the normal vector $(n + \lambda m)$ must be perpendicular to $l$,so $(n + \lambda m) \cdot l = 0$. $(n \cdot l) + \lambda(m \cdot l) = 0$. Calculating the dot products: $n \cdot l = (2)(2) + (-3)(-1) + (4)(1) = 4 + 3 + 4 = 11$. $m \cdot l = (1)(2) + (-1)(-1) + (0)(1) = 2 + 1 + 0 = 3$. Substituting these values: $11 + 3\lambda = 0 \Rightarrow \lambda = -\frac{11}{3}$. Now,substitute $\lambda$ back into the plane equation: $r \cdot ((2 \hat{i} - 3 \hat{j} + 4 \hat{k}) - \frac{11}{3}(\hat{i} - \hat{j})) = 1 - 4(-\frac{11}{3})$. $r \cdot ((\frac{6-11}{3}) \hat{i} + (\frac{-9+11}{3}) \hat{j} + 4 \hat{k}) = 1 + \frac{44}{3}$. $r \cdot (-\frac{5}{3} \hat{i} + \frac{2}{3} \hat{j} + 4 \hat{k}) = \frac{47}{3}$. Multiplying by $3$: $r \cdot (-5 \hat{i} + 2 \hat{j} + 12 \hat{k}) = 47$. Using $r = x \hat{i} + y \hat{j} + z \hat{k}$,we get $-5x + 2y + 12z = 47$,which simplifies to $5x - 2y - 12z + 47 = 0$.

If $n=2 \hat{i}-3 \hat{j}+4 \hat{k}$,$m=\hat{i}-\hat{j}$,and $l=2 \hat{i}-\hat{j}+\hat{k}$,then the Cartesian equation of the plane passing through the line of intersection of two planes $r \cdot n=1$ and $r \cdot m=-4$ and perpendicular to the plane $r \cdot l=-8$ is

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