Since c is orthogonal it must be parallel to their cross product:
c || a x b
Given:
a = i^ + 2j^ + 5k^
b = i^ + 2j^ - k^
The cross product:
a x b = | i^ j^ k^ |
| 1 2 5 |
|. 1 2 -1 |
$$= \hat{i}(2(-1) - 5 \cdot 2) - \hat{j}(1(-1) - 5 \cdot 1) + \hat{k}(1 \cdot 2 - 2 \cdot 1)$$
$$= \hat{i}(-2 - 10) - \hat{j}(-1 - 5) + \hat{k}(2 - 2)$$
$$= -12\hat{i} + 6\hat{j}$$
$$= 6(-2\hat{i} + \hat{j})$$
Hence $\vec{c}$ is parallel to $2\hat{i} - \hat{j}$.
