Let the shortest distance between the lines and be . Then the positive value of is [2025]

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Solution

Q.
Let the shortest distance between the lines $\frac{x - 3}{3} = \frac{y - α}{- 1} = \frac{z - 3}{1}$ and $\frac{x + 3}{- 3} = \frac{y + 7}{2} = \frac{z - β}{4}$ be $3 \sqrt{30}$ . Then the positive value of $5 α + β$ is [2025]

1 46

2 48

3 42

4 40

Ans.
(1)

Given line are $\frac{x - 3}{3} = \frac{y - α}{- 1} = \frac{z - 3}{1}$ and $\frac{x + 3}{- 3} = \frac{y + 7}{2} = \frac{z - β}{4}$

Let $A (3, α, 3)$ and $B (- 3, - 7, β)$

$∴ \vec{B A} = 6 \hat{i} + (α + 7) \hat{j} + (3 - β) \hat{k}$

Now, $\vec{p} \times \vec{q} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & - 1 & 1 \\ - 3 & 2 & 4 \end{matrix} |$ $= - 6 \hat{i} - 15 \hat{j} + 3 \hat{k}$

$∴$ Shortest distance between lines $=$ $| \frac{\vec{B A} \cdot (\vec{p} \times \vec{q})}{| \vec{p} \times \vec{q} |} |$ $= 3 \sqrt{3}$

$\Rightarrow 36 + 15 α + 105 - 9 + 3 β = 270 \Rightarrow 5 α + β = 46$ .

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