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
Show Hint
For problems involving lines and planes, remember the conditions for intersection, parallelism, and perpendicularity.
The scalar triple product is zero for coplanar vectors.
The shortest distance between a point and a line is along the perpendicular.
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The Correct Option is B
Solution and Explanation
Given:
\( (3x + 2y + z - 2) + \mu(x - 3y + 2z - 13) = 0 \)
Substituting values:
\( 3(3 + \mu) + 1 \cdot (2 - 3\mu) - 2(1 + 2\mu) = 0 \)
\( 9 - 4\mu = 0 \)
Solving for \( \mu \):
\( \mu = \frac{9}{4} \)
Next:
\( 4(-15 - 8 + \alpha - 2) + 9(-5 + 12 + 2\alpha - 13) = 0 \)
\( -100 + 4\alpha - 54 + 18\alpha = 0 \)
Simplifying:
\( \Rightarrow \alpha = 7 \)
Let:
\( P \equiv (3\lambda - 5, \lambda - 4, -2\lambda + 7) \)
Direction ratios of PQ:
\( (3\lambda - 1, \lambda - 1, -2\lambda + 5) \)
Since \( PQ \perp \ell_1 \):
\( 3(3\lambda - 1) + 1 \cdot (\lambda - 1) - 2(-2\lambda + 5) = 0 \)
Solving:
\( \lambda = 1 \)
Substituting \( \lambda = 1 \) into \( P \):
\( P = (-2, -3, 5) \)
Finally:
\( |a| + |b| + |c| = 10 \)
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