The distance between the planes $$x + 2y - 3z - 4 = 0$$     and $$2x + 4y - 6z = 1$$    along the line $$\frac{x}{1} = \frac{y}{{ - 3}} = \frac{z}{2}$$    is :

Solution :
Any point on $$\frac{x}{1} = \frac{y}{{ - 3}} = \frac{z}{2}$$   is $$\left( {r,\, - 3r,\,2r} \right).$$   It is on the first plane if
$$\eqalign{ & r + 2\left( { - 3r} \right) - 3\left( {2r} \right) - 4 = 0 \cr & {\text{or }} - 11r - 4 = 0 \cr & {\text{or }}r = - \frac{4}{{11}} \cr & \therefore {\text{ the point is }}\left( { - \frac{4}{{11}},\,\frac{{12}}{{11}},\, - \frac{8}{{11}}} \right) \cr & \left( {r,\, - 3r,\,2r} \right){\text{ is on the second plane if }} \cr & 2r + 4\left( { - 3r} \right) - 6\left( {2r} \right) = 1 \cr & {\text{or }} - 22r = 1 \cr & {\text{or }}r = - \frac{1}{{22}} \cr & \therefore {\text{ the point is }}\left( { - \frac{1}{{22}},\,\frac{3}{{22}},\, - \frac{1}{{11}}} \right) \cr & \therefore {\text{ The distance }} = \sqrt {{{\left( { - \frac{4}{{11}} + \frac{1}{{22}}} \right)}^2} + {{\left( {\frac{{12}}{{11}} - \frac{3}{{22}}} \right)}^2} + {{\left( { - \frac{8}{{11}} + \frac{1}{{11}}} \right)}^2}} \cr & = \sqrt {\frac{{49}}{{{{\left( {22} \right)}^2}}} + \frac{{441}}{{{{\left( {22} \right)}^2}}} + \frac{{49}}{{{{\left( {11} \right)}^2}}}} \cr & = \frac{1}{{22}}\sqrt {49 + 441 + 196} \cr} $$