I have
$$P=(1,-2,3)$$ And the x-axis with direction vector $(1,0,0)$ and y-axis with direction vector $(0,1,0)$.
By the angle formula:
$$\cos\theta = \frac{|\vec u\cdot\vec v|}{||\vec u||||\vec v||}$$
So I need to force the angle between the $x$ axis to be $45$ and the angle between the $y$ axis to be $60$. The problem is that when I consider the direction vector of my lineto be $(a,b,c)$ and I plug in the formula I don't know what to do next:
$$\cos 45= \frac{|(1,0,0)\cdot(a,b,c)|}{1||\sqrt{a^2+b^2+c^2}||}$$ and $$\cos 60= \frac{|(0,1,0)\cdot(a,b,c)|}{1||\sqrt{a^2+b^2+c^2}||}$$
I can't solve for $a,b,c$ and I don't know how to use the point in the formula.
HINT
You're on right track so far!
Next, let the angle with z-axis be $\alpha $, you get : $$\cos \alpha = \frac{|(0,0,1)\cdot(a,b,c)|}{1||\sqrt{a^2+b^2+c^2}||}$$
Squaring and adding these 3 equations you get : $$\cos^245 + \cos^260 + \cos^2\alpha = 1$$
Solve $\cos\alpha $ and you will have direction vector of the line