Key Concepts and Formulas

• Equation of a Plane in Intercept Form: The equation of a plane that passes through the x-intercept $(a, 0, 0)$, y-intercept $(0, b, 0)$, and z-intercept $(0, 0, c)$ is given by $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$.
• Fundamental Algebraic Identity: If $p+q+r=0$, then $p^3+q^3+r^3-3pqr=0$, which implies $p^3+q^3+r^3 = 3pqr$. This identity is crucial for simplifying expressions involving sums of cubes under a zero-sum condition.

Step-by-Step Solution

Step 1: Determine the Equation of Plane P

• What we are doing: We are finding the specific equation of the plane P using the given intercept points.
• Why we are doing it: The expression contains $x, y, z$ which are points on plane P. Knowing the relationship between $x, y, z$ from the plane's equation is fundamental to simplifying the expression.
• Calculation: The plane P passes through $(42, 0, 0)$, $(0, 42, 0)$, and $(0, 0, 42)$. These are the x, y, and z-intercepts, respectively, with $a=42$, $b=42$, and $c=42$. Using the intercept form of the plane equation: $\frac{x}{42} + \frac{y}{42} + \frac{z}{42} = 1$ Multiplying the entire equation by 42, we get the simplified equation of plane P: $x + y + z = 42$

Step 2: Transform the Plane Equation to Relate to the Expression's Terms

• What we are doing: We are manipulating the plane equation $x+y+z=42$ to create terms similar to those found in the given expression, i.e., $(x-11)$, $(y-19)$, and $(z-12)$.
• Why we are doing it: This step is critical to reveal a hidden algebraic condition that can be used for simplification. We observe that $11+19+12 = 42$.
• Reasoning: We can rewrite the constant $42$ as the sum of $11$, $19$, and $12$.
• Calculation: $x + y + z = 11 + 19 + 12$ Rearranging the terms by moving $11, 19, 12$ to the left side: $(x - 11) + (y - 19) + (z - 12) = 0$ This transformed equation provides a crucial relationship between the terms in the expression.

Step 3: Introduce Substitution for Simplification

• What we are doing: We are defining new, simpler variables to represent the complex binomial terms in the expression.
• Why we are doing it: Substituting complex terms with single variables makes the expression much cleaner and easier to recognize patterns or apply algebraic identities.
• Substitution: Let $a = x - 11$ Let $b = y - 19$ Let $c = z - 12$
• Consequence: From Step 2, we immediately have the condition: $a + b + c = 0$ This condition is central to applying the algebraic identity.

Step 4: Rewrite the Given Expression Using Substitutions

• What we are doing: We are replacing the original terms in the given expression with our new variables $a, b, c$. We also substitute the value of $x+y+z$ from the plane equation.
• Why we are doing it: This step converts the intimidating original expression into a much simpler form that can be algebraically manipulated.
• Original Expression: $3 + {{x - 11} \over {{{(y - 19)}^2}{{(z - 12)}^2}}} + {{y - 19} \over {{{(x - 11)}^2}{{(z - 12)}^2}}} + {{z - 12} \over {{{(x - 11)}^2}{{(y - 19)}^2}}} - {{x + y + z} \over {14(x - 11)(y - 19)(z - 12)}}$
• Substitution and Simplification: Substitute $a, b, c$ and $x+y+z=42$: $3 + \frac{a}{b^2c^2} + \frac{b}{a^2c^2} + \frac{c}{a^2b^2} - \frac{42}{14(a)(b)(c)}$ Simplify the last term: $\frac{42}{14abc} = \frac{3}{abc}$
• Rewritten Expression: $3 + \frac{a}{b^2c^2} + \frac{b}{a^2c^2} + \frac{c}{a^2b^2} - \frac{3}{abc}$

Step 5: Combine Fractional Terms and Apply the Algebraic Identity

• What we are doing: We are combining the fractional terms by finding a common denominator and then applying the algebraic identity $p^3+q^3+r^3-3pqr=0$ for $p=a, q=b, r=c$ since $a+b+c=0$.
• Why we are doing it: This is the final step to simplify the expression completely using the derived condition.
• Calculation: The common denominator for the fractional terms $\frac{a}{b^2c^2}$, $\frac{b}{a^2c^2}$, $\frac{c}{a^2b^2}$, and $\frac{3}{abc}$ is $a^2b^2c^2$. $3 + \frac{a \cdot a^2}{a^2b^2c^2} + \frac{b \cdot b^2}{a^2b^2c^2} + \frac{c \cdot c^2}{a^2b^2c^2} - \frac{3 \cdot abc}{a^2b^2c^2}$ $3 + \frac{a^3}{a^2b^2c^2} + \frac{b^3}{a^2b^2c^2} + \frac{c^3}{a^2b^2c^2} - \frac{3abc}{a^2b^2c^2}$ Combining the fractions: $3 + \frac{a^3 + b^3 + c^3 - 3abc}{a^2b^2c^2}$ Since we established $a+b+c=0$, we can apply the algebraic identity $a^3+b^3+c^3-3abc=0$. $3 + \frac{0}{a^2b^2c^2}$ Assuming $a, b, c \neq 0$ (which implies $x \neq 11, y \neq 19, z \neq 12$), the fraction becomes 0. $3 + 0 = 3$

Common Mistakes & Tips

• Missing the Intercept Form: Not recognizing the points $(42,0,0)$, $(0,42,0)$, $(0,0,42)$ as intercepts can lead to a longer method for finding the plane equation.
• Overlooking the Sum: The key insight of $11+19+12=42$ is crucial. Always look for numerical relationships between constants in the problem.
• Forgetting the Algebraic Identity: The identity $a+b+c=0 \implies a^3+b^3+c^3=3abc$ is a common tool in JEE. Mastering it saves significant time.
• Algebraic Errors in Combining Fractions: Be careful when finding common denominators and multiplying terms to avoid calculation mistakes.

Summary

This problem effectively tests the understanding of 3D plane equations and algebraic identities. The solution begins by determining the equation of the plane P, which is $x+y+z=42$. The critical step involves transforming this equation by observing that $11+19+12=42$, leading to $(x-11)+(y-19)+(z-12)=0$. By substituting $a=x-11$, $b=y-19$, and $c=z-12$, we obtain the condition $a+b+c=0$. The given complex expression is then rewritten in terms of $a, b, c$. Finally, by combining the fractional terms and applying the algebraic identity $a^3+b^3+c^3-3abc=0$ (valid because $a+b+c=0$), the entire expression simplifies to $3$.

The final answer is $\boxed{3}$, which corresponds to option (A).