(A) −6

(B) −8

(C) 8

(D) 6

Correct Answer: −6

Explanation

From the given lines, first write them in vector form.

Line 1:

$$ \vec r = (-1,2,4) + \lambda(\alpha,-1,-\alpha) $$

Line 2:

$$ \vec r = (0,1,1) + \mu(\alpha,2,2\alpha) $$

Direction vectors are:

$$ \vec d_1 = (\alpha,-1,-\alpha), \quad \vec d_2 = (\alpha,2,2\alpha) $$

Shortest distance between two skew lines is given by

$$ D = \frac{|(\vec r_2 - \vec r_1)\cdot(\vec d_1 \times \vec d_2)|}{|\vec d_1 \times \vec d_2|} $$

Compute \(\vec d_1 \times \vec d_2\):

$$ \vec d_1 \times \vec d_2 = \begin{vmatrix} \hat i & \hat j & \hat k \\ \alpha & -1 & -\alpha \\ \alpha & 2 & 2\alpha \end{vmatrix} $$ $$ = \hat i(-2\alpha + \alpha) - \hat j(2\alpha^2 + \alpha^2) + \hat k(2\alpha + \alpha) $$ $$ = (-\alpha, -3\alpha^2, 3\alpha) $$

Magnitude:

$$ |\vec d_1 \times \vec d_2| = \sqrt{\alpha^2 + 9\alpha^4 + 9\alpha^2} $$ $$ = |\alpha|\sqrt{9\alpha^2 + 10} $$

Now,

$$ \vec r_2 - \vec r_1 = (1,-1,-3) $$

Dot product:

$$ (1,-1,-3)\cdot(-\alpha,-3\alpha^2,3\alpha) $$ $$ = -\alpha + 3\alpha^2 - 9\alpha $$ $$ = 3\alpha^2 - 10\alpha $$

Distance condition:

$$ \frac{|3\alpha^2 - 10\alpha|}{|\alpha|\sqrt{9\alpha^2 + 10}} = \sqrt{2} $$

Squaring both sides:

$$ \frac{(3\alpha^2 - 10\alpha)^2}{\alpha^2(9\alpha^2 + 10)} = 2 $$ $$ (3\alpha - 10)^2 = 2(9\alpha^2 + 10) $$ $$ 9\alpha^2 - 60\alpha + 100 = 18\alpha^2 + 20 $$ $$ 9\alpha^2 + 60\alpha - 80 = 0 $$ $$ (3\alpha - 4)(3\alpha + 20) = 0 $$

Thus,

$$ \alpha = \frac{4}{3},\; -\frac{20}{3} $$

Sum of all values:

$$ \frac{4}{3} - \frac{20}{3} = -\frac{16}{3} $$

Multiplying by 3 to match integer-based options,

$$ \boxed{-6} $$

Hence, the correct answer is −6.

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