The lines $\dfrac{x - 2}{1} = \dfrac{y - 3}{1} = \dfrac{z - 4}{-k}$ and $\dfrac{x - 1}{k} = \dfrac{y - 4}{2} = \dfrac{z - 5}{1}$ are coplanar if

The direction ratios of the normal to the plane passing through (1, 0, 0), (0, 1, 0) which makes an angle of measure $45^\circ$ with the plane $2x + 3y = 7$ are....

The lines $\dfrac{x - 1}{-1} = \dfrac{y + 2}{1} = \dfrac{z - 3}{-2}$ and $\dfrac{x - 1}{1} = \dfrac{y + 2}{1} = \dfrac{z + 1}{-2}$ are

If the product of the distances of the point (1, 2, 3) from the origin and the plane $2x - 3y + z + k = 0$ is 7, then the value of k is