JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Three Dimensional Geometry

(x, y, z)

be an arbitrary point lying on a plane

P

which passes through the points

(42, 0, 0), (0, 42, 0)

and

(0, 0, 42)

, then the value of the expression

3 + \frac{x - 11}{( y - 19 ) ^{2} ( z - 12 ) ^{2}} + \frac{y - 9}{( x - 11 ) ^{2} ( z - 12 ) ^{2}} + \frac{z - 12}{( x - 11 ) ^{2} ( y - 19 ) ^{2}} - \frac{x + y + z}{14 ( x - 11 ) ( y - 19 ) ( z - 12 )}

is equal to :

Answer: (A)

Solution

The plane through

(42, 0, 0)

(0, 42, 0)

and

(0, 0, 42)

x + y + z = 42

. Put

u = x - 11

v = y - 19

and

w = z - 12

, so that

u + v + w = (x + y + z) - 42 = 0

For

u + v + w = 0

the identity

u^{3} + v^{3} + w^{3} = 3 uv w

holds, so the three fractions combine as

\frac{u}{v ^{2} w ^{2}} + \frac{v}{u ^{2} w ^{2}} + \frac{w}{u ^{2} v ^{2}} = \frac{u ^{3} + v ^{3} + w ^{3}}{u ^{2} v ^{2} w ^{2}} = \frac{3}{uv w}

The subtracted term is

\frac{x + y + z}{14 uv w} = \frac{42}{14 uv w} = \frac{3}{uv w}

, which cancels that sum exactly. The expression equals

3

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Single correct option