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

• A. 8
• B. 10
• C. 12
• D. 14

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. 

Answer 1

The Correct Option is B

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:

\( |a| + |b| + |c| = 10 \)