Given:

The points A (3, 4, 5) and B (-1, 3, -7)

x_{1} = 3, y_{1} = 4, z_{1} = 5;

x_{2} = -1, y_{2} = 3, z_{2} = -7;

PA^{2} + PB^{2} = k^{2} ……….(1)

Let the point be P (x, y, z).

Now by using distance formula,

We know that the distance between two points P (x_{1}, y_{1}, z_{1}) and Q (x_{2}, y_{2}, z_{2}) is given by

\( PQ= \sqrt{(x^2 – x^1)^2 + (y^2 – y^1)^2 + (z^2 – z^1)^2}\)

PA = \( \sqrt{(3 – x)^2 + (4 – y)^2 + (5 – z)^2}\)

PB = \( \sqrt{(-1 – x)^2 + (3 – y)^2 + (-7 – z)^2}\)

Now, substituting these values in (1), we have

[(3 – x)^{2} + (4 – y)^{2} + (5 – z)^{2}] +

[(-1 – x)^{2} + (3 – y)^{2} + (-7 – z)^{2}] = k^{2}

[(9 + x^{2} – 6x) + (16 + y^{2} – 8y) + (25 + z^{2} – 10z)] +

[(1 + x^{2} + 2x) + (9 + y^{2} – 6y) + (49 + z^{2} + 14z)] = k^{2}

9 + x^{2} – 6x + 16 + y^{2} – 8y + 25 + z^{2} – 10z + 1 +

x^{2} + 2x + 9 + y^{2} – 6y + 49 + z^{2} + 14z = k^{2}

2x^{2} + 2y^{2} + 2z^{2} – 4x – 14y + 4z + 109 = k^{2}

2x^{2} + 2y^{2} + 2z^{2} – 4x – 14y + 4z = k^{2} – 109

2 (x^{2} + y^{2} + z^{2} – 2x – 7y + 2z) = k^{2} – 109

(x^{2} + y^{2} + z^{2} – 2x – 7y + 2z) '

= \( \dfrac{ (k^2 – 109)}{2}\)

Hence, the required equation is (x^{2} + y^{2} + z^{2} – 2x – 7y + 2z)

= \( \dfrac{ (k^2 – 109)}{2}\)

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