The distance of the point (-1,2,3) from the plane r.→(i^−2j^+3k^)=10 parallel to the line of the shortest distance between the lines r→=(i^−j^)+λ(2i^+k^) and r→=(2i^−j^)+μ(i^−j^+k^) is

Line of shortest distance is parallel to i^j^k^2011−11=i^−j^−2k^ Line passing through A(−1,2,3) and having direction ratio (1,−1,−2) is x+11=y−2−1=z−3−2=λ Let the line intersect the plane at P P≡(λ−1,−λ+2,−2λ+3) P lies on the plane r.→(i^−2j^+3k^)=10 i.e. x−2y+3z=10 ∴(λ−1)−2(−λ+2)+3(−2λ+3)=10λ=−2P≡(−3,4,7)AP=4+4+16=26

The distance of the point (-1,2,3) from the plane r.→(i^−2j^+3k^)=10 parallel to the line of the shortest distance between the lines r→=(i^−j^)+λ(2i^+k^) and r→=(2i^−j^)+μ(i^−j^+k^) is

Line of shortest distance is parallel to i^j^k^2011−11=i^−j^−2k^ Line passing through A(−1,2,3) and having direction ratio (1,−1,−2) is x+11=y−2−1=z−3−2=λ Let the line intersect the plane at P P≡(λ−1,−λ+2,−2λ+3) P lies on the plane r.→(i^−2j^+3k^)=10 i.e. x−2y+3z=10 ∴(λ−1)−2(−λ+2)+3(−2λ+3)=10λ=−2P≡(−3,4,7)AP=4+4+16=26

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The distance of the point (-1,2,3) from the plane $\vec{r .} (\hat{i} - 2 \hat{j} + 3 \hat{k}) = 10$ parallel to the line of the shortest distance between the lines $\vec{r} = (\hat{i} - \hat{j}) + λ (2 \hat{i} + \hat{k})$ and $\vec{r} = (\hat{2 i} - \hat{j}) + μ (\hat{i} - \hat{j} + \hat{k})$ is

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$2 \sqrt{5}$

$3 \sqrt{6}$

$3 \sqrt{5}$

$2 \sqrt{6}$

answer is D.

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Detailed Solution

$Line of shortest distance is parallel to |\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & 1 \\ 1 & - 1 & 1 \end{array}| = \hat{i} - \hat{j} - 2 \hat{k} Line passing through A (- 1, 2, 3) and having direction ratio (1, - 1, - 2) is \frac{x + 1}{1} = \frac{y - 2}{- 1} = \frac{z - 3}{- 2} = λ Let the line intersect the plane at P P \equiv (λ - 1, - λ + 2, - 2 λ + 3) P lies on the plane \vec{r .} (\hat{i} - 2 \hat{j} + 3 \hat{k}) = 10 i . e . x - 2 y + 3 z = 10 \begin{array}{l} ∴ (λ - 1) - 2 (- λ + 2) + 3 (- 2 λ + 3) = 10 \\ λ = - 2 \\ P \equiv (- 3, 4, 7) \\ A P = \sqrt{4 + 4 + 16} = 2 \sqrt{6} \end{array}$