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: Understanding the Concept:
A plane that contains a line and a given point has a normal perpendicular to both the line direction and the vector from a point on the line to the given point.

Step 2: Collect the data.
The line has direction \(\bar{d} = (-3, 2, 1)\) and passes through \(A(2, 3, -2)\). The given point is \(B(3, 7, -7)\), so \(\overrightarrow{AB} = (1, 4, -5)\).

Step 3: Find the normal.
\[ \bar{n} = \bar{d}\times\overrightarrow{AB} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -3 & 2 & 1 \\ 1 & 4 & -5 \end{vmatrix} = (-14, -14, -14) \]
So the normal is along \((1, 1, 1)\).

Step 4: Equation of the plane.
The plane is \(x + y + z = c\). Put \(B(3, 7, -7)\): \(c = 3\). Check with \(A\): \(2 + 3 - 2 = 3\), which agrees.

Step 5: Distance from the origin.
\[ d = \frac{|3|}{\sqrt{3}} = \sqrt{3} \Rightarrow d^2 = 3 \]

Final Answer:
\(d^2 = 3\), option (B). \[ \boxed{3} \]