Two Iines L1:x=5,y3−α=z−2 and L2:x=α,y−1=z2−αare coplanar. Then, a can take value(s):

Then, x2−x1,y2−y1,z2−z1,a1,b1,c1and a2,b2,c2 are coplanar i.e., x2−x1y2−y1z2−z1a1b1c1a2b2c2=0Here, x=5,y3−α=z−2⇒ x−50=y−0−(α−3)=z−0−2and x=α,y−1=z2−α⇒ x−α0=y−0−1=z−02−α⇒ 5−α0003−α−20−12−α=0⇒ (5−α)[(3−α)(2−α)−2]=0⇒ (5−α)α2−5α+4=0⇒ (5−α)(α−1)(α−4)=0⇒ α=1,4,5

Two Iines L1:x=5,y3−α=z−2 and L2:x=α,y−1=z2−αare coplanar. Then, a can take value(s):

Then, x2−x1,y2−y1,z2−z1,a1,b1,c1and a2,b2,c2 are coplanar i.e., x2−x1y2−y1z2−z1a1b1c1a2b2c2=0Here, x=5,y3−α=z−2⇒ x−50=y−0−(α−3)=z−0−2and x=α,y−1=z2−α⇒ x−α0=y−0−1=z−02−α⇒ 5−α0003−α−20−12−α=0⇒ (5−α)[(3−α)(2−α)−2]=0⇒ (5−α)α2−5α+4=0⇒ (5−α)(α−1)(α−4)=0⇒ α=1,4,5

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Two Iines $L_{1} : x = 5, \frac{y}{3 - α} = \frac{z}{- 2}$ and $L_{2} : x = α, \frac{y}{- 1} = \frac{z}{2 - α}$
are coplanar. Then, a can take value(s):

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Detailed Solution

Then, $(x_{2} - x_{1}, y_{2} - y_{1}, z_{2} - z_{1}), (a_{1}, b_{1}, c_{1})$

and $(a_{2}, b_{2}, c_{2})$ are coplanar

$i.e., |\begin{array}{ccc} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{array}| = 0$

Here, $x = 5, \frac{y}{3 - α} = \frac{z}{- 2}$

$\Rightarrow \frac{x - 5}{0} = \frac{y - 0}{- (α - 3)} = \frac{z - 0}{- 2}$

and $x = α, \frac{y}{- 1} = \frac{z}{2 - α}$

$\begin{array}{l} \Rightarrow \frac{x - α}{0} = \frac{y - 0}{- 1} = \frac{z - 0}{2 - α} \\ \Rightarrow |\begin{array}{ccc} 5 - α & 0 & 0 \\ 0 & 3 - α & - 2 \\ 0 & - 1 & 2 - α \end{array}| = 0 \end{array}$

$\begin{array}{l} \Rightarrow (5 - α) [(3 - α) (2 - α) - 2] = 0 \\ \Rightarrow (5 - α) [α^{2} - 5 α + 4] = 0 \\ \Rightarrow (5 - α) (α - 1) (α - 4) = 0 \\ \Rightarrow α = 1, 4, 5 \end{array}$