Explanation

(c)
(1, 2, 3) satisfies the plane $x - 2y + \mathcal{z} = 0$ and also $(\widehat{i} + 2\widehat{j} + 3\widehat{k}) \bullet \left( \widehat{i} - 2\widehat{j} + \widehat{k} \right) = 0$
Since the lines $\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{\mathcal{z} - 3}{3}$ and $\frac{x}{1} = \frac{y}{2} = \frac{\mathcal{z}}{3}$ both satisfy (0, 0, 0) and (1, 2, 3), both are same. Given line is obviously parallel to the plane $x - 2y + \mathcal{z} = 6$