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(C) The position vector of the point $P(5, -4, -3)$ is $\vec{r} = 5\hat{i} - 4\hat{j} - 3\hat{k}$.The direction cosines of a vector $\vec{r} = a\hat{i

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$\frac{\pi}{4}$

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(C) The position vector of the point $P(5, -4, -3)$ is $\vec{r} = 5\hat{i} - 4\hat{j} - 3\hat{k}$.The direction cosines of a vector $\vec{r} = a\hat{i} + b\hat{j} + c\hat{k}$ are given by $\cos \alpha = \frac{a}{|\vec{r}|}$,$\cos \beta = \frac{b}{|\vec{r}|}$,and $\cos \gamma = \frac{c}{|\vec{r}|}$,where $\alpha, \beta, \gamma$ are the angles made with the $X, Y, Z$ axes respectively.First,calculate the magnitude of the vector $\vec{r}$:$|\vec{r}| = \sqrt{5^2 + (-4)^2 + (-3)^2} = \sqrt{25 + 16 + 9} = \sqrt{50} = 5\sqrt{2}$.The angle $\alpha$ with the positive $X$-axis is given by:$\cos \alpha = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}}$.Therefore,$\alpha = \cos^{-1}\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4}$.

The angle made by the position vector of the point $(5, -4, -3)$ with the positive direction of the $X$-axis is:

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