Question

Equation of the plane which passes through the point of intersection of lines $\frac{x-1}{3}=\frac{y-2}{1}=\frac{z-3}{2}$ and $\frac{x-3}{1}=\frac{y-1}{2}=\frac{z-2}{3}$ and at the greatest distance from the point (0, 0, 0) is

• A. $4x+3y+5z=25$
• B. $4x+3y+5z=50$
• C. $3x+4y+5z=59$
• D. $x+7y-5z=2$

Answer 1

The correct option is B

$4x+3y+5z=50$

Let a point $(3\lambda +1,\lambda +2,2\lambda +3)$ on the first line also lie on the second line. Then,
$\frac{3\lambda +1-3}{1}=\frac{\lambda +2-1}{2}=\frac{2\lambda +3-2}{3}\Rightarrow \lambda =1$
Hence the point of intersetion, P of the two lines is (4, 3, 5).
Since the plane passes through (4, 3, 5), the distance between the plane and the origin, O must at least be equal to OP.
The plane is furthest from the origin when OP is perpendicular to the plane. In that case, the direction ratios of the plane's normal are 4,3,5.
Equation of the plane perpendicular to OP and passing through P is 4x+ 3y+ 5z =50