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Mathematics Question on introduction to three dimensional geometry

Two lines L1:x=5 y3−α=z−2L_1:x = 5\, \frac{y}{3-\alpha} = \frac{z}{-2} and L1:x=α y−1=z2−αL_1:x = \alpha \, \frac{y}{-1} =\frac{z}{2-\alpha} are coplaner. Then, α\alpha can take value(s)

A

1

B

2

C

3

D

4

Answer

4

Explanation

Solution

If two straight lines are coplanar,
i.e,    x−x1a1=y−y1b1=z−z1c1\ \ \ \frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}
and    x−x2a2=y−y2b2=z−z2c2\ \ \ \frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2} are coplanar
Then, (x2−x1,y2−y1,z2−z1),(a1,b1,c1)(x_2 - x_1,y_2- y_1,z_2 -z_1),(a_1,b_1,c_1) and (a2,b2,c2)(a_2,b_2,c_2) are
coplanar,
i.e. \hspace10mm ∣x2−x1y2−y1z2−z1 a1b1c1 a1b2c2 ∣=0 \begin{vmatrix} x_2 -x_1 & y_2 - y_1 & z_2 - z_1 \\\ a_1 & b_1 & c_1 \\\ a_1 & b_2 & c_2 \\\ \end{vmatrix}= 0
Here, \hspace20mm x =5, \frac{y}{3-\alpha} = \frac{z}{-2}
\Rightarrow \hspace20mm \frac{x - 5}{0} =\frac{y-0}{-(-\alpha-3)} = \frac{z-0}{-2} \hspace20mm ...(i)
and \hspace20mm x = \alpha, \frac{y}{-1} = \frac{z}{2- \alpha}
⇒\Rightarrow \hspace20mm \frac{x - \alpha}{0} =\frac{y-0}{-1} = \frac{z-0}{2- \alpha} \hspace25mm ...(ii)
⇒\Rightarrow \hspace10mm ∣5−α00 03−α−2 0−12−α ∣=0 \begin{vmatrix} 5 -\alpha & 0 & 0 \\\ 0 & 3- \alpha & -2 \\\ 0 & -1 & 2-\alpha \\\ \end{vmatrix}= 0
⇒     (5−α)[(3−α) (2−α)−2]=0\Rightarrow \ \ \ \ \ (5-\alpha) [(3- \alpha)\ (2- \alpha) -2] = 0
\Rightarrow \hspace15mm (5-\alpha) [ \alpha^2-5\alpha+4] = 0
\Rightarrow \hspace15mm (5-\alpha) (\alpha - 1)(\alpha-4) = 0
\therefore \hspace50mm \alpha = 1,4,5