The distance of the point (1, 3, -7) from the plane passing through the point (1, -1, -1), having normal perpendicular to both the lines $\frac{x-1}{1}=\frac{y+2}{-2}=\frac{z-4}{3}$ and $\frac{x-2}{2}=\frac{y+1}{-1}=\frac{z+7}{-1}$, is
Answer & explanation
Correct answer: option 2
Let the equation of the plane through the point (1, -1, -1) be
$a(x-1) + b(y+1)+c (z +1) = 0 $ ......(i)
The direction ratios of the normal to the plane are proportional to a, b, c. The normal is perpendicular to the lines
$\frac{x-1}{1}=\frac{y+2}{-2}=\frac{z-4}{3}$ and $\frac{x-2}{2}=\frac{y+1}{-1}=\frac{z+7}{-1}$.
$∴ a - 2b + 3c = 0 $ and, $ 2a - b - c= 0 $
Using cross-multiplication, we obtain
$\frac{a}{5}=\frac{b}{7}=\frac{c}{3}$
Substituting the values of a, b, c in (i), we obtain
$5(x-1) + 7(y+1) + 3(z+1)= 0 $ or, $5x + 7y + 3z + 5 = 0 $....(ii)
The distance 'd' of the point (1, 3, -7) from plane (ii) is
$d= \begin{vmatrix}\frac{5+21-21+5}{\sqrt{25+49+9}}\end{vmatrix}=\frac{10}{\sqrt{83}}$