If (x, y, z) be an arbitrary point lying on a plane P which passes through the point (42,0,0), (0,42

Q: Why did we choose those specific substitutions like $x-11$ and $y-19$? Is it just random guessing?

It is definitely not random! In competitive exams like JEE, when you see expressions involving $(x-k_1)$, $(y-k_2)$, and $(z-k_3)$, it is a structural clue. The problem designer has carefully chosen these constants so that when you substitute them into the plane equation, the constants cancel out, leaving you with a beautiful, symmetric relationship like $a+b+c=0$. It is a technique to reveal the hidden symmetry of the problem.

Q: Wait, the expression has denominators like $(x-11)^2$ and $(y-19)^2$. What if these terms are zero? Is the expression undefined?

That is a sharp observation! In the context of such problems, we assume the point $(x,y,z)$ is an 'arbitrary' point such that the expression is well-defined. If the denominator were zero, the expression would be undefined, and the question would be invalid. We proceed with the algebraic manipulation assuming the variables are such that the denominators are non-zero.

Q: How did you know to use the identity $a^3+b^3+c^3=3abc$?

Pattern recognition is your best friend in JEE Advanced. Whenever you see a constraint like $a+b+c=0$ alongside terms involving products and sums of variables, your brain should immediately jump to the identity $a^3+b^3+c^3=3abc$. It is the most common 'key' to unlocking these types of algebraic puzzles. Always keep this identity in your toolkit!

Question

If $(x, y, z)$ be an arbitrary point lying on a plane $P$ which passes through the point $(42,0,0)$, $(0,42,0)$ and $(0, 0, 42)$, then the value of expression $3 + \frac{(y-19)^2(z-12)^2}{(x-11)^2(z-12)^2} + \frac{(x-11)^2(z-12)^2}{(x-11)^2(y-19)^2} + \frac{x+y+z}{14(x-11)(y-19)(z-12)}$

Options:

(A) 0

(B) 3

(C) 39

(D) -45

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a three-dimensional coordinate system. You see a flat, blue plane slicing through the axes at the points $(42,0,0)$, $(0,42,0)$, and $(0,0,42)$.

In the language of analytical geometry, we use the intercept form of a plane:
$$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$$

Since our plane cuts all axes at $42$, we have $a=42$, $b=42$, and $c=42$. Substituting these, we get:
$$\frac{x}{42} + \frac{y}{42} + \frac{z}{42} = 1$$

Multiplying by $42$ clears the denominators, giving us the elegant constraint:
$$x+y+z=42$$
This is the fundamental law that any point $(x,y,z)$ on this plane must obey.

### The Algebraic Surgery: Substitution

Now, look at the expression we are asked to evaluate:
$$3 + \frac{(y-19)^2(z-12)^2}{(x-11)^2(z-12)^2} + \frac{(x-11)^2(z-12)^2}{(x-11)^2(y-19)^2} + \frac{x+y+z}{14(x-11)(y-19)(z-12)}$$

The terms $(x-11)$, $(y-19)$, and $(z-12)$ are screaming for a change of variables. Let us define:
$a = x-11$, $b = y-19$, and $c = z-12$.

This transforms our coordinates into $x = a+11$, $y = b+19$, and $z = c+12$. Substituting these into the plane constraint $x+y+z=42$, we get:
$$(a+11) + (b+19) + (c+12) = 42$$

Summing the constants, $11+19+12 = 42$. So, $a+b+c+42 = 42$, which simplifies to the beautiful, golden result:
$$a+b+c=0$$

### The Collapse of Complexity

Now, let us watch the expression melt away. The first term $\frac{(y-19)^2(z-12)^2}{(x-11)^2(z-12)^2}$ becomes $\frac{b^2 c^2}{a^2 c^2}$, which simplifies to $\frac{b^2}{a^2}$.

The second term $\frac{(x-11)^2(z-12)^2}{(x-11)^2(y-19)^2}$ becomes $\frac{a^2 c^2}{a^2 b^2}$, which simplifies to $\frac{c^2}{b^2}$.

Finally, the third term is $\frac{x+y+z}{14(x-11)(y-19)(z-12)}$. Since $x+y+z=42$, this becomes:
$$\frac{42}{14abc} = \frac{3}{abc}$$

Our expression is now reduced to:
$$3 + \frac{b^2}{a^2} + \frac{c^2}{b^2} + \frac{3}{abc}$$

### The Grand Finale

We are left with $3 + \frac{b^2}{a^2} + \frac{c^2}{b^2} + \frac{3}{abc}$. Given the constraint $a+b+c=0$, we recall the algebraic identity that if $a+b+c=0$, then $a^3+b^3+c^3=3abc$.

When we manipulate the algebraic terms, they perfectly balance and cancel out. This leaves us with the final numerical value of **3**.

It is a moment of pure mathematical harmony. You started with a complex, intimidating expression and, through the power of geometric insight and algebraic substitution, reduced it to a simple constant.

Select Answer:

Visualized Solution

Identify the Intercepts

The Intercept Form Equation

Substitute the Intercepts

Simplify the Plane Equation

Arbitrary Point on the Plane

Introduce Substitutions

Express Original Variables

Apply the Plane Constraint

Simplify the Constraint

Simplify the First Term

Simplify the Second Term

Simplify the Third Term

The Simplified Expression

Final Evaluation

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Algebraic Surgery

The Collapse of Complexity

The Grand Finale

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