JEE Main 2024 — 4 April, Shift 1 — Question 29

$$\text{Given:\; } \triangle ABC\; \text{ with\; area\; } 15\sqrt{2},\; \text{ and\; vectors:}$$ $$\vec{AB} = \mathbf{i}+2\mathbf{j}-7\mathbf{k},\quad \vec{AC} = 6\mathbf{i}+d\mathbf{j}-2\mathbf{k},\quad \vec{BC} = a\mathbf{i}+b\mathbf{j}+c\mathbf{k}$$

$\text{Step\; 1:\; Use\; area\; formula:}$

$$\text{Area\;} = \frac{1}{2}\|\vec{AB} \times \vec{AC}\| \implies \|\vec{AB} \times \vec{AC}\| = 30\sqrt{2}$$

$\text{Step\; 2:\; Compute\; cross\; product:}$

$$\vec{AB} \times \vec{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & -7 \\ 6 & d & -2 \end{vmatrix} = (7d-4)\mathbf{i} - 40 \mathbf{j} + (d-12)\mathbf{k}$$

$\text{Step\; 3:\; Magnitude\; squared: }$

$$\|\vec{AB} \times \vec{AC}\|^2 = (7d-4)^2 + (-40)^2 + (d-12)^2 = 1800$$

$\text{Solve\; for\; } d>0:$

$$5d^2 - 8d - 4 = 0 \implies d = 2$$

$\text{Step\; 4:\; Compute\; vector\; } \vec{BC}:$

$$\vec{BC} = \vec{AC}-\vec{AB} = (6-1, 2-2, -2-(-7)) = (5,0,5)$$

$\text{Step\; 5:\; Compute\; squares\; of\; side\; lengths: }$

$$|\vec{AB}|^2 = 1^2 + 2^2 + (-7)^2 = 54$$ $$|\vec{AC}|^2 = 6^2 + 2^2 + (-2)^2 = 44$$ $$|\vec{BC}|^2 = 5^2 + 0^2 + 5^2 = 50$$

$\text{Step\; 6:\; Identify\; the\; largest\; side: }$

$$\max(|\vec{AB}|^2, |\vec{AC}|^2, |\vec{BC}|^2) = |\vec{AB}|^2 = 54$$ $$\boxed{54}$$