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(A) Let the given point be $A = \hat{i} - \hat{j} + 2\hat{k}$ and the direction vector of the line be $\vec{v} = 3\hat{i} + \hat{j} + \hat{k}$.The equ

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$10\hat{i} + 2\hat{j} + 5\hat{k}$

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(A) Let the given point be $A = \hat{i} - \hat{j} + 2\hat{k}$ and the direction vector of the line be $\vec{v} = 3\hat{i} + \hat{j} + \hat{k}$.The equation of the line passing through $A$ and parallel to $\vec{v}$ is $\vec{r} = A + \lambda\vec{v}$.Any point $P$ on this line is given by $P = (\hat{i} - \hat{j} + 2\hat{k}) + \lambda(3\hat{i} + \hat{j} + \hat{k})$.The vector $\vec{AP} = P - A = \lambda(3\hat{i} + \hat{j} + \hat{k})$.The distance between $A$ and $P$ is $|\vec{AP}| = |\lambda| \sqrt{3^2 + 1^2 + 1^2} = |\lambda| \sqrt{11}$.Given that the distance is $3\sqrt{11}$,we have $|\lambda| \sqrt{11} = 3\sqrt{11}$,which implies $|\lambda| = 3$,so $\lambda = \pm 3$.For $\lambda = 3$,$P = (\hat{i} - \hat{j} + 2\hat{k}) + 3(3\hat{i} + \hat{j} + \hat{k}) = (1+9)\hat{i} + (-1+3)\hat{j} + (2+3)\hat{k} = 10\hat{i} + 2\hat{j} + 5\hat{k}$.For $\lambda = -3$,$P = (\hat{i} - \hat{j} + 2\hat{k}) - 3(3\hat{i} + \hat{j} + \hat{k}) = (1-9)\hat{i} + (-1-3)\hat{j} + (2-3)\hat{k} = -8\hat{i} - 4\hat{j} - \hat{k}$.Thus,the possible position vectors are $10\hat{i} + 2\hat{j} + 5\hat{k}$ and $-8\hat{i} - 4\hat{j} - \hat{k}$.

Find the position vector of the point on the line passing through the point $\hat{i} - \hat{j} + 2\hat{k}$ and parallel to the vector $3\hat{i} + \hat{j} + \hat{k}$,which is at a distance of $3\sqrt{11}$ units from the point $\hat{i} - \hat{j} + 2\hat{k}$.

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