Question

The distance of the point (1, –2, 3) from the plane x-y+z=5 measured parallel to the line $\frac{x}{2}=\frac{y}{3}=\frac{z}{-6}$, is

• A. 1
• B. $\frac{6}{7}$
• C. $\frac{7}{6}$
• D. 7

Answer 1

The correct option is A 1

Direction cosines of line =$(\frac{2}{7},\frac{3}{7},\frac{-6}{7})$
Now, $x^{\prime }=1+\frac{2r}{7},y^{\prime }=-2+\frac{3r}{7}$ and $z^{\prime }=3-\frac{6r}{7}$
$\therefore (1+\frac{2r}{7})-(-2+\frac{3r}{7})+(3-\frac{6r}{7})=5\Rightarrow r=1$.