A survey laser is aimed from the origin of a coordinate system with the z-axis pointing up…

Step 1. Two cosines. $l = \cos 60^\circ = \dfrac{1}{2}$ and $m = \cos 45^\circ = \dfrac{\sqrt{2}}{2}$.

Step 2. The third. $n^2 = 1 - \dfrac{1}{4} - \dfrac{1}{2} = \dfrac{1}{4}$, so $n = \pm\dfrac{1}{2}$.

Step 3. The angles. $n = \dfrac{1}{2}$ gives $\gamma = 60^\circ$ for the rising beam. $n = -\dfrac{1}{2}$ gives $\gamma = 120^\circ$ for the falling beam.

Step 4. The vector. For the rising beam, $10\left(\dfrac{1}{2}\hat i + \dfrac{\sqrt{2}}{2}\hat j + \dfrac{1}{2}\hat k\right) = 5\hat i + 5\sqrt{2}\hat j + 5\hat k$. The $y$-component is $5\sqrt{2} \approx 7.07$ metres.