JEE MAIN - Mathematics (2022 - 26th July Morning Shift - No. 21)
Let $$\mathrm{Q}$$ and $$\mathrm{R}$$ be two points on the line $$\frac{x+1}{2}=\frac{y+2}{3}=\frac{z-1}{2}$$ at a distance $$\sqrt{26}$$ from the point $$P(4,2,7)$$. Then the square of the area of the triangle $$P Q R$$ is ___________.
Answer
153
Explanation
$$L:{{x + 1} \over 2} = {{y + 2} \over 3} = {{2 - 1} \over 2}$$
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Let $$T(2t - 1,\,3t - 2,\,2t + 1)$$
$$\because$$ $$PT\,{ \bot ^r}\,QR$$
$$\therefore$$ $$2(2t - 5) + 3(3t - 4) + 2(2t - 6) = 0$$
$$17t = 34$$
$$\therefore$$ $$t = 2$$
So $$T(3,4,5)$$
$$\therefore$$ $$PT = \sqrt {1 + 4 + 4} = 3$$
$$\therefore$$ $$QT = \sqrt {26 - 9} = \sqrt {17} $$
$$\therefore$$ Area of $$\Delta PQR = {1 \over 2} \times 2\sqrt {17} \times 3 = 3\sqrt {17} $$
$$\therefore$$ Square of $$ar(\Delta PQR) = 153$$.
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