If for some $m\in R$ the lines $L_1:\frac{x+1}{m} = \frac{y-m}{-1} = \frac{z-1}{1}$ and $L_2:\frac{x+2}{-4} = \frac{y+1}{9} = \frac{z+1}{1}$ are coplanar, then line $L_1$ passes through the point

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: Parametrise L1 for m = 2
Points on $L_1$: $(-1+2t,\ 2-t,\ 1+t)$.

Step 2: Find the point
Set $t=4$: $(7,-2,5)$, option (B). This uses $m=2$, one of the two values that make the lines coplanar.

Final Answer:
Point $(7,-2,5)$ lies on $L_1$ for $m=2$, option (B). \[ \boxed{(7,-2,5)} \]

If for some \(m\in R\) the lines \(L_1:\frac{x+1}{m} = \frac{y-m}{-1} = \frac{z-1}{1}\) and \(L_2:\frac{x+2}{-4} = \frac{y+1}{9} = \frac{z+1}{1}\) are coplanar, then line \(L_1\) passes through the point

Show Hint

The Correct Option is B

Solution and Explanation

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