If the line vec{r} = (hat{i} - 2hat{j} - hat{ | vedclass

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(A) line with direction vector $\vec{b} = 2\hat{i} + \hat{j} + 2\hat{k}$ is parallel to a plane with normal vector $\vec{n} = 3\hat{i} - 2\hat{j} - m\

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$2$

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(A) line with direction vector $\vec{b} = 2\hat{i} + \hat{j} + 2\hat{k}$ is parallel to a plane with normal vector $\vec{n} = 3\hat{i} - 2\hat{j} - m\hat{k}$ if and only if the line is perpendicular to the normal of the plane.This implies that the dot product of the direction vector of the line and the normal vector of the plane must be zero,i.e.,$\vec{b} \cdot \vec{n} = 0$.Substituting the vectors,we get: $(2\hat{i} + \hat{j} + 2\hat{k}) \cdot (3\hat{i} - 2\hat{j} - m\hat{k}) = 0$.Calculating the dot product: $(2)(3) + (1)(-2) + (2)(-m) = 0$.$6 - 2 - 2m = 0$.$4 - 2m = 0$.$2m = 4$.$m = 2$.

If the line $\vec{r} = (\hat{i} - 2\hat{j} - \hat{k}) + \lambda (2\hat{i} + \hat{j} + 2\hat{k})$ is parallel to the plane $\vec{r} \cdot (3\hat{i} - 2\hat{j} - m\hat{k}) = 14$,then the value of $m$ is:

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