Three Dimensional Geometry — Class 12 Maths Solution

Equation of the two planes are $x + 2y = 0$ and $3y - z = 0$.

Let $\overrightarrow {{{\rm{n}}_1}}$ and ${\overrightarrow {\rm{n}} _2}$ are the normals to the two planes, respectively.

$\therefore$ ${\overrightarrow {\rm{n}} _1} = \widehat {\rm{i}} + 2\widehat {\rm{j}}$and $\overrightarrow {{{\rm{n}}_2}} = 3\widehat {\rm{j}} - \widehat {\rm{k}}$

Since, required line is parallel to the given two planes. Therefore,

$\overrightarrow {\rm{b}} = {\overrightarrow {\rm{n}} _1} \times {\overrightarrow {\rm{n}} _2} = \left| {\begin{array}{cccccccccccccccccccc}{\widehat {\rm{i}}}&{\widehat {\rm{j}}}&{\widehat {\rm{k}}}\\1&2&0\\0&3&{ - 1}\end{array}} \right|$

$= \widehat {\rm{i}}( - 2) - \widehat {\rm{j}}( - 1) + \widehat {\rm{k}}(3)$
$= - 2\widehat {\rm{i}} + \widehat {\rm{j}} + 3\widehat {\rm{k}}$

So, the equation of the lines through the point (3,0,1) and parallel to the given two planes are

$(x - 3)\widehat {\rm{i}} + (y - 0)\widehat {\rm{j}} + (z - 1)\widehat {\rm{k}} + \lambda ( - 2\widehat {\rm{i}} + \widehat {\rm{j}} + 3\widehat {\rm{k}})$

$\Rightarrow$ $(x - 3)\widehat {\rm{i}} + y\widehat {\rm{j}} + (z - 1)\widehat {\rm{k}} + \lambda ( - 2\widehat {\rm{i}} + \widehat {\rm{j}} + 3\widehat {\rm{k}})$