Step 1: Understanding the Question:
We need to find the equation of a plane that passes through a given point and contains a line defined by two other points.
Once the equation is found, we identify the intercepts \( p, q, r \) and calculate their sum.
Step 2: Key Formula or Approach:
A plane containing a line joining points \( A(x_1, y_1, z_1) \) and \( B(x_2, y_2, z_2) \) and passing through point \( P(x_3, y_3, z_3) \) has a normal vector \( \vec{n} = \vec{PA} \times \vec{PB} \).
Step 3: Detailed Explanation:
Let the given points be \( P(2, 1, -1) \), \( A(1, 3, 2) \), and \( B(1, 2, 1) \).
Vector \( \vec{PA} = (1-2)\hat{i} + (3-1)\hat{j} + (2-(-1))\hat{k} = -\hat{i} + 2\hat{j} + 3\hat{k} \).
Vector \( \vec{PB} = (1-2)\hat{i} + (2-1)\hat{j} + (1-(-1))\hat{k} = -\hat{i} + \hat{j} + 2\hat{k} \).
The normal to the plane is \( \vec{n} = \vec{PA} \times \vec{PB} \):
\[ \vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
-1 & 2 & 3
-1 & 1 & 2 \end{vmatrix} \]
\[ \vec{n} = \hat{i}(4-3) - \hat{j}(-2-(-3)) + \hat{k}(-1-(-2)) = 1\hat{i} - 1\hat{j} + 1\hat{k} \]
The equation of the plane passing through \( P(2, 1, -1) \) with normal \( (1, -1, 1) \) is:
\[ 1(x-2) - 1(y-1) + 1(z-(-1)) = 0 \]
\[ x - 2 - y + 1 + z + 1 = 0 \implies x - y + z = 0 \]
This plane passes through the origin.
For a plane passing through the origin, all three intercepts on the axes are 0.
Thus, \( p = 0, q = 0, r = 0 \).
The sum \( p + q + r = 0 + 0 + 0 = 0 \).
Step 4: Final Answer:
The sum of the intercepts is 0.