Q67 / 125 · BITSAT 2015 · Mathematics
Two lines $L_{1}: x=5, \frac{y}{3-\alpha}=\frac{z}{-2}$
$\mathrm{L}_{2}: \mathrm{x}=\alpha, \frac{\mathrm{y}}{-1}=\frac{\mathrm{z}}{2-\alpha}$ are coplanar. Then, $\alpha \mathrm{can}$ take value (s)
- Correct answerA1,4,5
- B1,2,5
- C3,4,5
- D2,4,5
Solution
The equations of given lines can be written
as
$$
\begin{array}{l}
\mathrm{L}_{1}: \mathrm{x}-5=\frac{\mathrm{y}}{3-\alpha}=\frac{\mathrm{z}}{-2} \\
\mathrm{~L}_{2}: \mathrm{x}-\alpha=\frac{\mathrm{y}}{-1}=\frac{\mathrm{z}}{2-\alpha}
\end{array}
$$
Since, these lines are coplanar.
$$
\begin{array}{l}
\text { Therefore, }\left|\begin{array}{ccc}
5-\alpha & 0-0 & 0-0 \\
0 & 3-\alpha & -2 \\
0 & -1 & 2-\alpha
\end{array}\right|=0 \\
\Rightarrow \quad(5-\alpha)(3-\alpha)(2-\alpha)-2=0 \\
\Rightarrow \quad(5-\alpha)\left(6-3 \alpha-2 \alpha+\alpha^{2}-2\right]=0 \\
\Rightarrow \quad(5-\alpha)\left(\alpha^{2}-5 \alpha+4\right]=0 \\
\Rightarrow \quad(5-\alpha)(\alpha-1)(\alpha-4)=0 \\
\Rightarrow \quad \alpha=1,4,5
\end{array}
$$
Hence, the correct answer is option: 'A'.