If the lines $x = -1+s$, $y = 3-λs$, $z = 1+λs$ and $x = \frac{t}{2}$, $y = 1+t$, $z = 2-t$ with parameters s and t, are coplanar, then $λ =$

Question

Let the lines l₁: $\frac{x+5}{3} = \frac{y+4}{1}=\frac{z−α}{−2}$ and l₂: 3x + 2y + z – 2 = 0 = x – 3y + 2z - 13 be coplanar. If the point P(a, b, c) on l₁ is nearest to the point Q(- 4, -3, 2), then |a| + |b|+|c| is equal to

Answer 1

Step 1: Convert
Line 2 is $\frac{x}{1/2} = \frac{y - 1}{1} = \frac{z - 2}{-1}$, direction proportional to $(1, 2, -2)$.

Step 2: Triple product
The scalar triple product of $\overrightarrow{PQ} = (1, -2, 1)$, $(1, -\lambda, \lambda)$ and $(1, 2, -2)$ equals $-2 - \lambda$.

Step 3: Condition
Zero gives $\lambda = -2$. Option (C).

Final Answer:
Option (C). \[ \boxed{-2} \]

If the lines \(x = -1+s\), \(y = 3-λs\), \(z = 1+λs\) and \(x = \frac{t}{2}\), \(y = 1+t\), \(z = 2-t\) with parameters s and t, are coplanar, then \(λ =\)

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The Correct Option is C

Solution and Explanation

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