Question

Let a plane P pass through the point \((3,7,-7)\) and contain the line \(\frac{x-2}{-3} = \frac{y-3}{2} = \frac{z+2}{1}\). If the distance of the plane P from the origin is d, then \(d^2\) is

• A. \(2\)
• B. \(3\)
• C. \(4\)
• D. \(6\)

Hint:

Find the normal as the cross product of the line direction and a vector joining a point on the line to the given point.

Answer 1

The Correct Option is B

Step 1: Use a general plane through the line.
A plane containing the line can be written with normal $\bar{n} = (a, b, c)$ where $-3a + 2b + c = 0$, and through point $(2, 3, -2)$.

Step 2: Impose the second point.
The plane passes through $(3, 7, -7)$, so $a(1) + b(4) + c(-5) = 0$.

Step 3: Solve for the ratios.
From the first, $c = 3a - 2b$. Put into the second: $a + 4b - 15a + 10b = 0$, so $14b = 14a$ and $b = a$. Then $c = a$. The normal is $(1, 1, 1)$.

Step 4: Plane and distance.
The plane is $x + y + z = 3$. Its distance from the origin is $3/\sqrt{3}$, so $d^2 = 9/3 = 3$.

Final Answer:
Option (B). \[ \boxed{3} \]