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 $a + b + c = (x - 11) + (y - 19) + (z - 12) = x + y + z - 42 = 42 - 42 = 0$.Substituting these into the expression:$E = 3 + \frac{a}{b^2 c^2} + \frac{b}{a^2 c^2} + \frac{c}{a^2 b^2} - \frac{x + y + z}{14abc}$Since $x + y + z = 42$,the last term is $-\frac{42}{14abc} = -\frac{3}{abc}$.$E = 3 + \frac{a^3 + b^3 + c^3}{a^2 b^2 c^2} - \frac{3}{abc} = 3 + \frac{a^3 + b^3 + c^3 - 3abc}{a^2 b^2 c^2}$.Since $a + b + c = 0$,we know that $a^3 + b^3 + c^3 = 3abc$.Therefore,$E = 3 + \frac{3abc - 3abc}{a^2 b^2 c^2} = 3 + 0 = 3$.

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