If (x, y, z) is an arbitrary point lying on a | vedclass

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Question

(B) The equation of a plane passing through $(42, 0, 0)$,$(0, 42, 0)$,and $(0, 0, 42)$ in intercept form is $\frac{x}{42} + \frac{y}{42} + \frac{z}{42

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) The equation of a plane passing through $(42, 0, 0)$,$(0, 42, 0)$,and $(0, 0, 42)$ in intercept form is $\frac{x}{42} + \frac{y}{42} + \frac{z}{42} = 1$,which simplifies to $x + y + z = 42$.We can rewrite this as $(x-11) + (y-19) + (z-12) = 42 - 11 - 19 - 12 = 0$.Let $a = x-11$,$b = y-19$,and $c = z-12$. Then $a + b + c = 0$.The given expression is $3 + \frac{a}{b^2 c^2} + \frac{b}{a^2 c^2} + \frac{c}{a^2 b^2} - \frac{42}{14abc}$.Since $a+b+c=0$,we have $x+y+z=42$. Substituting this into the last term,we get $\frac{42}{14abc} = \frac{3}{abc}$.The expression becomes $3 + \frac{a^3 + b^3 + c^3}{a^2 b^2 c^2} - \frac{3}{abc} = 3 + \frac{a^3 + b^3 + c^3 - 3abc}{a^2 b^2 c^2}$.Since $a+b+c=0$,the identity $a^3 + b^3 + c^3 = 3abc$ holds.Therefore,the expression simplifies to $3 + \frac{3abc - 3abc}{a^2 b^2 c^2} = 3 + 0 = 3$.

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