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(D) Given that point $A$ has position vector $\vec{a} = \hat{i}-\hat{j}+3 \hat{k}$.The line is parallel to the vector $\vec{v} = 2 \hat{i}+\hat{j}-2 \

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$13 \hat{i}+5 \hat{j}-9 \hat{k}$

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(D) Given that point $A$ has position vector $\vec{a} = \hat{i}-\hat{j}+3 \hat{k}$.The line is parallel to the vector $\vec{v} = 2 \hat{i}+\hat{j}-2 \hat{k}$.The unit vector in the direction of the line is $\hat{u} = \frac{\vec{v}}{|\vec{v}|} = \frac{2 \hat{i}+\hat{j}-2 \hat{k}}{\sqrt{2^2+1^2+(-2)^2}} = \frac{2 \hat{i}+\hat{j}-2 \hat{k}}{3}$.Since $P$ lies on the line and $|AP|=18$,the vector $\vec{AP}$ can be written as $\vec{AP} = \pm 18 \hat{u} = \pm 18 \left( \frac{2 \hat{i}+\hat{j}-2 \hat{k}}{3} \right) = \pm 6(2 \hat{i}+\hat{j}-2 \hat{k}) = \pm (12 \hat{i}+6 \hat{j}-12 \hat{k})$.Using the triangle law of vector addition,the position vector of $P$ is $\vec{OP} = \vec{OA} + \vec{AP}$.Case $1$: $\vec{OP} = (\hat{i}-\hat{j}+3 \hat{k}) + (12 \hat{i}+6 \hat{j}-12 \hat{k}) = 13 \hat{i}+5 \hat{j}-9 \hat{k}$.Case $2$: $\vec{OP} = (\hat{i}-\hat{j}+3 \hat{k}) - (12 \hat{i}+6 \hat{j}-12 \hat{k}) = -11 \hat{i}-7 \hat{j}+15 \hat{k}$.Comparing with the given options,$13 \hat{i}+5 \hat{j}-9 \hat{k}$ is present.

If $P$ is a point lying on the line passing through the point $A(\hat{i}-\hat{j}+3 \hat{k})$ and parallel to the vector $2 \hat{i}+\hat{j}-2 \hat{k}$ such that $|AP|=18$,then a position vector of $P$ is

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