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

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(B) The equation of the plane passing through $(42, 0, 0)$,$(0, 42, 0)$,and $(0, 0, 42)$ is given by the intercept form: $\frac{x}{42} + \frac{y}{42}

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) The equation of the plane passing through $(42, 0, 0)$,$(0, 42, 0)$,and $(0, 0, 42)$ is given by the intercept form: $\frac{x}{42} + \frac{y}{42} + \frac{z}{42} = 1$,which simplifies to $x + y + z = 42$. Let $a = x-11$,$b = y-19$,and $c = z-12$. Then $a+b+c = (x+y+z) - (11+19+12) = 42 - 42 = 0$. We need to evaluate the expression $E = 3 + \frac{a}{b^2c^2} + \frac{b}{a^2c^2} + \frac{c}{a^2b^2} - \frac{x+y+z}{14abc}$. Since $x+y+z = 42$,the last term is $-\frac{42}{14abc} = -\frac{3}{abc}$. The expression becomes $E = 3 + \frac{a^3+b^3+c^3}{a^2b^2c^2} - \frac{3}{abc}$. Since $a+b+c=0$,we know that $a^3+b^3+c^3 = 3abc$. Substituting this into the expression: $E = 3 + \frac{3abc}{a^2b^2c^2} - \frac{3}{abc} = 3 + \frac{3}{abc} - \frac{3}{abc} = 3$.

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