Correct Answer - Option 1 : x – 2y + z = 0
Calculation:
Let the direction ratios of the plane containing lines \(\frac{x}{3} = \frac{y}{4} = \frac{z}{2}{\rm{\;and\;}}\frac{x}{4} = \frac{y}{2} = \frac{z}{3}{\rm{\;is}}\left\langle {\hat i,\hat j,\hat k} \right\rangle\)
The plane normal to the plane containing two given lines is:
\(\Rightarrow {n_1} = \left| {\begin{array}{*{20}{c}}
{\hat i}&{\hat j}&{\hat k}\\
3&4&2\\
4&2&3
\end{array}} \right|\)
\(\Rightarrow {n_1} = \hat i\left( {12 - 4} \right) - \hat j\left( {9 - 8} \right) + \hat k\left( {6 - 16} \right)\)
\(\Rightarrow {n_1} = \hat i\left( 8 \right) - \hat j\left( 1 \right) + \hat k\left( { - 10} \right)\)
From question, the plane containing above two lines and another plane is perpendicular to each other.
\(\Rightarrow \left( {8\hat i, - 1\hat j, - 10\hat k} \right) \times \left( {2\hat i + 3\hat j + 4\hat k} \right)\)
\(\Rightarrow \left| {\begin{array}{*{20}{c}}
{\hat i}&{\hat j}&{\hat k}\\
8&{ - 1}&{ - 10}\\
2&3&4
\end{array}} \right| = 0\)
\(\Rightarrow \hat i\left( { - 4 + 30} \right) - \hat j\left( {32 + 20} \right) + \hat k\left( {24 + 2} \right) = 0\)
\(\Rightarrow \hat i\left( {26} \right) - \hat j\left( {52} \right) + \hat k\left( {26} \right) = 0\)
\(\therefore \left( {\hat i - 2\hat j + \hat k} \right) = 0\)
Therefore, equation of plane is x - 2y + z = 0
