$A (Z_{1}), B (Z_{2}), C (Z_{3})$ vertices of $Δ A B C, | Z_{1} - Z_{2} | = 10, | Z_{2} - Z_{3} | = 17$ , $| Z_{3} - Z_{1} | = 21$ .point $P (Z_{4})$ satisfies $|2 Z - Z_{1} - Z_{2}| = 10$ then the greatest possible absolute value of $|\begin{matrix} z_{1} & \bar{z_{1}} & 1 \\ z_{3} & \bar{z_{3}} & 1 \\ z_{4} & \bar{z_{4}} & 1 \end{matrix}|$ equals

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answer is 378.

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Detailed Solution

$A B = 10, B C = 17, C A = 21$

P lies on circle with $\bar{A B}$ as diameter to find greatest area of $(Δ A P C) \times 4$
P should be farthest from AC, on the circle .
Let M be the mid point of AB (in center of circle)
Q be foot from M on AC ,now $P Q = P M + M Q = \frac{1}{2} (A B) + \frac{1}{2} (h_{B})$ (where $h_{B}$ = length of altitude from B to AC)
But area of $Δ A B C = \sqrt{24.14.7.3} = 84$
So, $h_{B} = \frac{2 (84)}{A C} (∵ area = \frac{1}{2} (A C) (h_{B}))$

$= \frac{2 \times 84}{21} = 8$

$∴ P Q = \frac{1}{2} (10) + \frac{1}{2} (8) = 5 + 4 = 9 {(G E)}_{greatest} = 4 Δ A B C = 4 (\frac{1}{2} \times 2 \times 9) = 2 \times 21 \times 9 = 378$