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**Hint:**Let $O$ be the origin and $P$ be any point. If $OP = r$ and $a,b,c$ be the Direction Ratios of$OP$ , then Direction Cosines of $OP$ are $ \pm \dfrac{a}{r}, \pm \dfrac{b}{r}, \pm \dfrac{c}{r}.$

Let $O$ be the origin and $P(x,y,z)$ be any point. Also if $OP = r$ and $l,m,n$ be Direction Cosines of $OP$ then $x = lr,y = mr,z = nr.$ so the coordinates of P are $(lr,mr,nr).$

**Complete step-by-step answer:**

It is given that $OP = 6$, and also given that their direction ratios as -2, 4,-4.

From the given hint and the given Direction Ratios of $OP$ Also $OP = 6$

The Direction Cosines of $OP$ are $ \pm \left( {\dfrac{{ - 2}}{6}} \right), \pm \left( {\dfrac{4}{6}} \right), \pm \left( {\dfrac{{ - 4}}{6}} \right)$ .

Now let us simplify the direction cosines we get, $ \pm \left( {\dfrac{{ - 1}}{3}} \right), \pm \left( {\dfrac{2}{3}} \right), \pm \left( {\dfrac{{ - 2}}{3}} \right)$

Now let us take positive signs in the Direction Cosines of $OP$.

Hence we get the following values $\dfrac{{ - 1}}{3},\dfrac{2}{3},\dfrac{{ - 2}}{3}.$

From $ \pm \left( {\dfrac{{ - 1}}{3}} \right), \pm \left( {\dfrac{2}{3}} \right), \pm \left( {\dfrac{{ - 2}}{3}} \right)$ this value let us consider the negative sign in the Direction Cosines of $OP$ $\dfrac{1}{3},\dfrac{{ - 2}}{3},\dfrac{2}{3}.$

Let the coordinates of P be $(x,y,z).$

Now let us take the following values $\dfrac{{ - 1}}{3},\dfrac{2}{3},\dfrac{{ - 2}}{3}.$

Hence we get, $l = \dfrac{{ - 1}}{3},\quad m = \dfrac{2}{3},\quad n = \dfrac{{ - 2}}{3}$ and $r = 6$ we get,

From the given hint we can find the value of x, y, z $x = \dfrac{{ - 1}}{3} \times 6 = - 2,\;\;y = \dfrac{2}{3} \times 6 = 4,\;\;z = \dfrac{{ - 2}}{3} \times 6 = - 4.$

Here x=-2, y=4 and z=-4.

So the coordinates of $P$ are $\left( { - 2,4, - 4} \right).$

Now let us consider the following values $\dfrac{1}{3},\dfrac{{ - 2}}{3},\dfrac{2}{3}.$

Hence we get, $l = \dfrac{1}{3},\quad m = \dfrac{{ - 2}}{3},\quad n = \dfrac{2}{3}$ and $r = 6$ we get,

Let us solve using the hint for x,y,z $x = \dfrac{1}{3} \times 6 = 2,\;\;y = \dfrac{{ - 2}}{3} \times 6 = - 4,\;\;z = \dfrac{2}{3} \times 6 = 4.$

Here x=2, y=-4 and z=4.

In this case the coordinates of $P$ are $\left( {2, - 4,4} \right).$

Hence,

We have found that the coordinates of $P$ are either $A)\;\;2,\; - 4,\;4$ or $B)\;\; - 2,\;4,\; - 4.$

**Note:**In the process of finding Direction Cosines from Direction Ratios the same sign must be taken throughout. Sign should not be changed.

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