Question

A plane meets the \(X, Y, Z\)-axes in \(A, B, C\) respectively. If the centroid of the triangle \(ABC\) is \((2,-3,5)\), then the perpendicular distance from origin to the given plane is

• A. \(\frac{7}{\sqrt{40}}\)
• B. \(\frac{6}{7}\)
• C. \(\frac{8}{\sqrt{50}}\)
• D. \(\frac{90}{19}\)

Hint:

If a plane cuts the axes at \((a,0,0)\), \((0,b,0)\), and \((0,0,c)\), then its equation is \[ \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1. \] The centroid of these intercept points is \[ \left(\frac{a}{3},\frac{b}{3},\frac{c}{3}\right). \]

Answer 1

The Correct Option is D

Step 1: Find intercepts from the centroid.
Centroid of triangle with intercepts \((a,0,0),(0,b,0),(0,0,c)\) is \((a/3, b/3, c/3)=(2,-3,5)\). So \(a=6, b=-9, c=15\).

Step 2: Write the plane equation.
\(\dfrac{x}{6}+\dfrac{y}{-9}+\dfrac{z}{15}=1 \Rightarrow 15x-10y+6z=90\).

Step 3: Find the perpendicular distance from the origin.
\[d = \frac{|90|}{\sqrt{15^2+10^2+6^2}} = \frac{90}{\sqrt{225+100+36}} = \frac{90}{\sqrt{361}} = \frac{90}{19}\]
\[ \boxed{\dfrac{90}{19}} \]