test

[Broyojo]

Step 1: Let's solve this problem step by step.

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• Bad
• Error Realization

Step 2: Given that $a^{2}+b^{2}+c^{2}=14$ and $d^{2}+e^{2}+f^{2}=77$, we have two sums of squares.

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• Error Realization

Step 3: We also have the given product of the respective terms $ad + be + cf = 32$.

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• Error Realization

Step 4: We are asked to find the value of $(bf-ce)^2 + (cd-af)^2 + (ae-bd)^2$.

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• Error Realization

Step 5: Now, we can use the identity $(x-y)^2 = x^2 - 2xy + y^2$.

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• Error Realization

Step 6: So let's expand each of the terms:

[
(bf-ce)^2 = b^2f^2 - 2bfce + c^2e^2,
]
[
(cd-af)^2 = c^2d^2 - 2cdfa + a^2f^2,
]
[
(ae-bd)^2 = a^2e^2 - 2aebd + b^2d^2.

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• Error Realization

Step 7: ]

Summing these up, we get:
[
(bf-ce)^2 + (cd-af)^2 + (ae-bd)^2 = b^2f^2 + c^2e^2 + c^2d^2 + a^2f^2 + a^2e^2 + b^2d^2 - 2(bfce + cdfa + aebd).
Step 8: ]

Notice that $b^2f^2 + c^2e^2 + a^2f^2 + c^2d^2 + a^2e^2 + b^2d^2$ can be rearranged to group the $a^2$, $b^2$, and $c^2$ terms together:

[
a^2(f^2 + e^2) + b^2(d^2 + f^2) + c^2(d^2 + e^2).
Step 9: ]

Given that $a^2 + b^2 + c^2 = 14$ and $d^2 + e^2 + f^2 = 77$, we can substitute $f^2 + e^2$ with $77 - d^2$ and $d^2 + f^2$ with $77 - e^2$ and $d^2 + e^2$ with $77 - f^2$:

[
a^2(77 - d^2) + b^2(77 - e^2) + c^2(77 - f^2).
Step 10: ]

Now, multiply each term out:

[
77a^2 - a^2d^2 + 77b^2 - b^2e^2 + 77c^2 - c^2f^2.
Step 11: ]

Since $a^2 + b^2 + c^2 = 14$, we can multiply this sum by 77:

[
77(a^2 + b^2 + c^2) - (a^2d^2 + b^2e^2 + c^2f^2).
Step 12: ]

Substitute $14$ for $a^2 + b^2 + c^2$:

[
77 \cdot 14 - (a^2d^2 + b^2e^2 + c^2f^2).
Step 13: ]

Now, we have to deal with the term $-2(bfce + cdfa + aebd)$.
Step 14: Note that this term is actually $-2(ad + be + cf)(af + bd + ce)$.
Step 15: Since $ad + be + cf = 32$, we need to find $af + bd + ce$ to compute this product.
Step 16: However, we do not have a direct way to find $af + bd + ce$.
Step 17: But, if we realize that the quantity we want to find, $(bf-ce)^2 + (cd-af)^2 + (ae-bd)^2$, can be written as a sum of squares, we can use the identity:

[
(a^2 + b^2 + c^2)(d^2 + e^2 + f^2) = (ad + be + cf)^2 + (bf - ce)^2 + (cd - af)^2 + (ae - bd)^2.
Step 18: ]

Substituting the given values:

[
14 \cdot 77 = 32^2 + (bf - ce)^2 + (cd - af)^2 + (ae - bd)^2.
Step 19: ]

We can now solve for $(bf - ce)^2 + (cd - af)^2 + (ae - bd)^2$:

[
(bf - ce)^2 + (cd - af)^2 + (ae - bd)^2 = 14 \cdot 77 - 32^2.
Step 20: ]

Calculating the numerical values:

[
(bf - ce)^2 + (cd - af)^2 + (ae - bd)^2 = 1078 - 1024.
Step 21: ]

[
(bf - ce)^2 + (cd - af)^2 + (ae - bd)^2 = 54.
Step 22: ]

The final answer is $\boxed{54}$.