The perpendicular distance from the origin to the plane containing the two lines,
$\frac{x + 2}{3} = \frac{y - 2}{5} = \frac{z + 5}{7}$ and $\frac{x - 1}{1} = \frac{y - 4}{4} = \frac{z + 4}{7}$ is :

\frac{11}{\sqrt{6}}

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6 \sqrt{11}

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11

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11 \sqrt{6}

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Solution

The correct option is B $\frac{11}{\sqrt{6}}$
$∣ ∣ ∣ ∣ \begin{matrix} i & j & k 3 & 5 & 7 1 & 4 & 7 \end{matrix} ∣ ∣ ∣ ∣$
$^i (35 - 28) -^j (21.7) +^k (12 - 5)$
$7^i - 14^j + 7^k$
$^i - 2^j +^k$
$1 (x + 2) - 2 (y - 2) + 1 (z + 15) = 0$
$x - 2 y + z + 11 = 0$
$\frac{11}{\sqrt{4 + 1 + 1}} = \frac{11}{\sqrt{6}}$

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Similar questions

\frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1}

\frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1}

\frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1}

\frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1}

\frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1} and \frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1} .

\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} and \frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 5}{5}

\frac{x - 1}{2} = \frac{y + 1}{3} = z and \frac{x + 1}{3} = \frac{y - 2}{1}; z = 2

\frac{x - 1}{- 1} = \frac{y + 2}{1} = \frac{z - 3}{- 2} and \frac{x - 1}{1} = \frac{y + 1}{2} = \frac{z + 1}{- 2}

\frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1} and \frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1}

\frac{x - 1}{- 3} = \frac{y - 2}{2 λ} = \frac{z - 3}{2} and \frac{x - 1}{3 λ} = \frac{y - 1}{1} = \frac{z - 6}{- 5}

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