Question

If \((3,4,-7)\) is the foot of the perpendicular drawn from the point \((-2,3,6)\) to the plane \(\pi\), then the sum of the intercepts made by the plane \(\pi\) on the \(x\)- and \(y\)-axes is

• A. \(132\)
• B. \(142\)
• C. \(210\)
• D. \(175\)

Hint:

If a point is the foot of perpendicular from another point to a plane, then the line joining the point and its foot is normal to the plane.

Answer 1

The Correct Option is A

Step 1: Understand the geometry.
The perpendicular from $A(-2,3,6)$ hits the plane at $P(3,4,-7)$. Because $AP$ is perpendicular to the plane, the vector $\overrightarrow{AP}$ is the normal vector to the plane.
Step 2: Find the normal vector.
\[\overrightarrow{AP} = (3-(-2),\ 4-3,\ -7-6) = (5,\ 1,\ -13)\] So the normal to plane $\pi$ is $\mathbf{n} = (5, 1, -13)$.
Step 3: Write the equation of the plane.
The plane passes through $P(3,4,-7)$ with normal $(5,1,-13)$. Using the point-normal form: \[5(x-3) + 1(y-4) - 13(z+7) = 0\] \[5x - 15 + y - 4 - 13z - 91 = 0\] \[5x + y - 13z = 110\]
Step 4: Find the $x$-intercept.
Set $y = 0$ and $z = 0$: \[5x = 110 \implies x = 22\] The $x$-intercept is $22$.
Step 5: Find the $y$-intercept.
Set $x = 0$ and $z = 0$: \[y = 110\] The $y$-intercept is $110$.
Step 6: Compute the sum of intercepts.
Sum $= 22 + 110 = 132$.
Step 7: State the final answer.
\[ \boxed{132} \]