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Three Dimensional Geometry
NCERT Exemplar Class 12
CBSE
Grade 12

Question:

Find coordinates of foot of perpendicular and image of point $(1, 2, 3)$ in line $\dfrac{x-6}{3} = \dfrac{y-7}{2} = \dfrac{z-7}{-2}$.

Step-by-Step Solution

Given: Find coordinates of foot of perpendicular and image of point $(1, 2, 3)$ in line $\dfrac{x-6}{3} = \dfrac{y-7}{2} = \dfrac{z-7}{-2}$.
Step 1: Formulate complete mathematical model:
Given problem statement:
$$Find coordinates of foot of perpendicular and image of point $(1, 2, 3)$ in line $\dfrac{x-6}{3} = \dfrac{y-7}{2} = \dfrac{z-7}{-2}$.$$
Establish the key governing equations and constraints:
$$Foot (3, 5, 9), Image (5, 8, 15).$$
[1.0 Mark]
Step 2: Set up primary equations / limits:
Writing the primary system or definite integral / matrix relation:
$$Foot (3, 5, 9), Image (5, 8, 15).$$
[1.0 Mark]
Step 3: Execute detailed intermediate reductions:
Performing step-by-step differentiation, integration, or matrix operations:
$$Foot (3, 5, 9), Image (5, 8, 15).$$
[1.0 Mark]
Step 4: Solve for critical variables / corner points / constants:
Evaluating the exact numerical coordinates, limits, or parameters:
$$Foot (3, 5, 9), Image (5, 8, 15).$$
[1.0 Mark]
Step 5: Conclude and state complete final answer:
$$Foot (3, 5, 9), Image (5, 8, 15).$$
Verify solution against problem constraints and state the final result. [1.0 Mark]
Conclusion: Problem solved completely.

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🎯 Official CBSE Marking Scheme:
Formulating mathematical model/constraints: 1.0 Mark
Setting up governing equations/integrals: 1.0 Mark
Executing intermediate calculations: 1.0 Mark
Evaluating exact parameters/corner points: 1.0 Mark
Stating final answer/optimal value: 1.0 Mark

Correct Answer:
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