Question

The distance of the point of intersection of the line $\frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2}$ and the plane x+y+z =17 from the point (3, 4, 5) is given by

• A. 3
• B. $\frac{3}{2}$
• C. $\sqrt{3}$
• D. $5$

Answer 1

The correct option is A 3

Any point on the line $\frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2}$ is (r+3, 2r+4, 2r+5) which satisfies the plane.
So, $r+3+2r+4+2r+5=17\Rightarrow r=1$.
$\therefore$ The point is (4, 6, 7).
Hence required distance is $\sqrt{1^2+2^2+2^2}=3$.