Let the shortest distance between the lines $\frac{x-3}{3}=\frac{y-\alpha}{-1}=\frac{z-3}{1}$ and $\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-\beta}{4}$ be $3 \sqrt{30}$. Then the positive value of $5 \alpha+\beta$ is

JEE Main · 2025 · 04 Apr 2025 · Shift 1 · Three Dimensional Geometry

Let the shortest distance between the lines

\frac{x - 3}{3} = \frac{y - α}{- 1} = \frac{z - 3}{1}

and

\frac{x + 3}{- 3} = \frac{y + 7}{2} = \frac{z - β}{4}

330

. Then the positive value of

5 α + β

(A)42
(B)46✓ Correct
(C)48
(D)40

Answer: (B)

Solution

With directions

p = (3, - 1, 1)

and

q = (- 3, 2, 4)

and points

A = (3, α, 3)

B = (- 3, - 7, β)

, the cross product is

p \times q = (- 6, - 15, 3)

, whose magnitude is

270 = 330

. With

A B = (- 6, - 7 - α, β - 3)

, the shortest distance is

\frac{∣ A B \cdot ( p \times q ) ∣}{3 30}

, and

A B \cdot (p \times q) = 132 + 15 α + 3 β

. Setting the distance to

330

gives

∣132 + 15 α + 3 β ∣ = 270

; the branch giving a positive result is

15 α + 3 β = 138

, so

5 α + β = 46

JEE Main · 2025 · 04 Apr 2025 · Shift 1 · Single correct option

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