If d1 is the shortest distance between the lines x + 1 = 2y = –12z, x = y + 2 = 6z – 6 and d2 is the shortest distance between the lines $\frac{\mathrm{x}-<!--\; -\; -->1}{2}=\frac{\mathrm{y}+8}{-<!--\; -\; -->7}=\frac{\mathrm{z}-<!--\; -\; -->4}{5},\frac{\mathrm{x}-<!--\; -\; -->1}{2}=\frac{\mathrm{y}-<!--\; -\; -->2}{1}=\frac{\mathrm{z}-<!--\; -\; -->6}{-<!--\; -\; -->3}$,  then the value of $\frac{32\sqrt{3}{\mathrm{d}}_1}{{\mathrm{d}}_2}$ is _____

The correct answer is 16.

We need to find d₁ (shortest distance between the first pair of lines) and d₂ (shortest distance between the second pair), then evaluate $\frac{32\sqrt{3}{d}_1}{d_2}$.

The formula for shortest distance between two skew lines with points a₁, a₂ and direction vectors b₁, b₂ is:

$d=\frac{\left|\right({\mathbf{a}}_2-{\mathbf{a}}_1)·({\mathbf{b}}_1\times {\mathbf{b}}_2\left)\right|}{|{\mathbf{b}}_1\times {\mathbf{b}}_2|}$

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STEP 1 — Finding d₁
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Rewrite Line 1: x + 1 = 2y = -12z in standard form:

$\frac{x+1}{1}=\frac{y}{1/2}=\frac{z}{-1/12}$

Point: a₁ = (-1, 0, 0)  |  Direction (scaled by 12): b₁ = (12, 6, -1)

Rewrite Line 2: x = y + 2 = 6z - 6 in standard form:

$\frac{x}{1}=\frac{y+2}{1}=\frac{z-1}{1/6}$

Point: a₂ = (0, -2, 1)  |  Direction (scaled by 6): b₂ = (6, 6, 1)

Compute a₂ - a₁ = (0-(-1), -2-0, 1-0) = (1, -2, 1)

Compute b₁ × b₂:

${\mathbf{b}}_1\times {\mathbf{b}}_2=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 12& 6& -1\\ 6& 6& 1\end{array}\right|$

= i(6·1 - (-1)·6) - j(12·1 - (-1)·6) + k(12·6 - 6·6)
= i(6 + 6) - j(12 + 6) + k(72 - 36)
= (12, -18, 36)

Magnitude: $|{\mathbf{b}}_1\times {\mathbf{b}}_2|=\sqrt{{12}^2+{18}^2+{36}^2}=\sqrt{144+324+1296}=\sqrt{1764}=42$

Dot product: (a₂ - a₁) · (b₁ × b₂) = (1)(12) + (-2)(-18) + (1)(36) = 12 + 36 + 36 = 84

$d_1=\frac{84}{42}=2$

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STEP 2 — Finding d₂
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Line 3: $\frac{x-1}{2}=\frac{y+8}{-7}=\frac{z-4}{5}$

Point: a₃ = (1, -8, 4)  |  Direction: b₃ = (2, -7, 5)

Line 4: $\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-6}{-3}$

Point: a₄ = (1, 2, 6)  |  Direction: b₄ = (2, 1, -3)

Compute a₄ - a₃ = (1-1, 2-(-8), 6-4) = (0, 10, 2)

Compute b₃ × b₄:

${\mathbf{b}}_3\times {\mathbf{b}}_4=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 2& -7& 5\\ 2& 1& -3\end{array}\right|$

= i((-7)(-3) - 5·1) - j(2·(-3) - 5·2) + k(2·1 - (-7)·2)
= i(21 - 5) - j(-6 - 10) + k(2 + 14)
= (16, 16, 16)

Magnitude: $|{\mathbf{b}}_3\times {\mathbf{b}}_4|=\sqrt{{16}^2+{16}^2+{16}^2}=\sqrt{768}=16\sqrt{3}$

Dot product: (a₄ - a₃) · (b₃ × b₄) = (0)(16) + (10)(16) + (2)(16) = 0 + 160 + 32 = 192

$d_2=\frac{192}{16\sqrt{3}}=\frac{12}{\sqrt{3}}=4\sqrt{3}$

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STEP 3 — Final Calculation
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Now substitute d₁ = 2 and d₂ = 4√3 into the expression:

$\frac{32\sqrt{3}·d_1}{d_2}=\frac{32\sqrt{3}·2}{4\sqrt{3}}=\frac{64\sqrt{3}}{4\sqrt{3}}=\frac{64}{4}=16$

The √3 terms cancel perfectly, giving the clean integer result. Therefore, the value of $\frac{32\sqrt{3}{d}_1}{d_2}$ is 16.