Question

Let the plane passing through point (2, 1, -1) containing line joining the points (1, 3, 2) and (1, 2, 1) makes intercepts \( p, q, r \) on co-ordinate axes, then \( p + q + r = \)}

• A. 0
• B. 3
• C. 2
• D. -2

Hint:

If a plane passes through the origin, its equation has no constant term ($d=0$).

Answer 1

The Correct Option is B

Step 1: Find Normal to Plane
Points: $A(2,1,-1), B(1,3,2), C(1,2,1)$.
Vectors in plane: $\vec{AB} = (-1, 2, 3)$, $\vec{AC} = (-1, 1, 2)$.
Normal $\vec{n} = \vec{AB} \times \vec{AC} = \hat{i} - \hat{j} + \hat{k}$.
Step 2: Equation of Plane
$1(x-2) - 1(y-1) + 1(z+1) = 0 \implies x - y + z = 0$.
Step 3: Analyze Intercepts
For $x - y + z = 0$, the plane passes through the origin. Intercepts are not defined in the standard form $x/p + y/q + z/r = 1$. If the equation was different (e.g., $x+y+z=c$), we would sum the intercepts.
Re-check point $C(1,2,1)$: $1-2+1=0$ (True). So the plane is $x-y+z=0$. Sum of intercepts for a plane passing through origin is usually considered 0 or undefined.
Step 4: Conclusion
Based on the calculation $x-y+z=0$.
Final Answer:(A)