Let P(3, 2, 6) be a point in space and Q be a | vedclass

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(A) Any point $Q$ on the line $\vec{r}$ can be represented as $Q(1 - 3\mu, -1 + \mu, 2 + 5\mu)$.The vector $\overrightarrow{PQ}$ is given by $\overrig

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$\frac{1}{4}$

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(A) Any point $Q$ on the line $\vec{r}$ can be represented as $Q(1 - 3\mu, -1 + \mu, 2 + 5\mu)$.The vector $\overrightarrow{PQ}$ is given by $\overrightarrow{PQ} = (1 - 3\mu - 3)\hat{i} + (-1 + \mu - 2)\hat{j} + (2 + 5\mu - 6)\hat{k} = (-3\mu - 2)\hat{i} + (\mu - 3)\hat{j} + (5\mu - 4)\hat{k}$.Since $\overrightarrow{PQ}$ is parallel to the plane $x - 4y + 3z = 1$,it must be perpendicular to the normal vector $\vec{n} = \hat{i} - 4\hat{j} + 3\hat{k}$ of the plane.Therefore,$\overrightarrow{PQ} \cdot \vec{n} = 0$.$1(-3\mu - 2) - 4(\mu - 3) + 3(5\mu - 4) = 0$.$-3\mu - 2 - 4\mu + 12 + 15\mu - 12 = 0$.$8\mu - 2 = 0$.$8\mu = 2$,which gives $\mu = \frac{1}{4}$.

Let $P(3, 2, 6)$ be a point in space and $Q$ be a point on the line $\vec{r} = (\hat{i} - \hat{j} + 2\hat{k}) + \mu(-3\hat{i} + \hat{j} + 5\hat{k})$. Then the value of $\mu$ for which the vector $\overrightarrow{PQ}$ is parallel to the plane $x - 4y + 3z = 1$ is

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