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(D) Let $l, m, n$ be the direction cosines of vector $r$. Since the vector is equally inclined with the coordinate axes,we have $l = m = n$.We know th

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$2\sqrt{3}(i + j + k)$

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(D) Let $l, m, n$ be the direction cosines of vector $r$. Since the vector is equally inclined with the coordinate axes,we have $l = m = n$.We know that $l^2 + m^2 + n^2 = 1$. Substituting $l = m = n$,we get $3l^2 = 1$,which implies $l^2 = \frac{1}{3}$.Since the tip of $r$ is in the positive octant,$l, m, n$ must be positive. Thus,$l = m = n = \frac{1}{\sqrt{3}}$.The vector $r$ is given by $r = |r|(li + mj + nk)$.Given $|r| = 6$,we have $r = 6 \left( \frac{1}{\sqrt{3}}i + \frac{1}{\sqrt{3}}j + \frac{1}{\sqrt{3}}k \right)$.Simplifying this,$r = \frac{6}{\sqrt{3}}(i + j + k) = 2\sqrt{3}(i + j + k)$.

$A$ vector $r$ is equally inclined with the coordinate axes. If the tip of $r$ is in the positive octant and $|r| = 6$,then $r$ is

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