Let the vertices Q and R of the triangle PQR lie on the line , QR = 5 and the coordinates of the point P be (0, 2, 3). If the area of the triangle PQR is then: [2025]

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Solution

Q.
Let the vertices Q and R of the triangle PQR lie on the line $\frac{x + 3}{5} = \frac{y - 1}{2} = \frac{z + 4}{3}$ , QR = 5 and the coordinates of the point P be (0, 2, 3). If the area of the triangle PQR is $\frac{m}{n}$ then: [2025]

1 $m - 5 \sqrt{21} n = 0$

2 $5 m - 2 \sqrt{21} n = 0$

3 $2 m - 5 \sqrt{21} n = 0$

4 $5 m - 21 \sqrt{2} n = 0$

Ans.
(3)

We have, equation of line L is

$\frac{x + 3}{5} = \frac{y - 1}{2} = \frac{z + 4}{3} = λ (say)$

$∴$ Coordinates of $M (5 λ - 3, 2 λ + 1, 3 λ - 4)$

Dr's of PM are $< 5 λ - 3, 2 λ - 1, 3 λ - 7 >$

Dr's of line L are < 5, 2, 3 >

As PM $⊥$ L, so, $(5 λ - 3) 5 + (2 λ - 1) 2 + (3 λ - 7) 3 = 0$

$\Rightarrow λ = 1$

$∴$ Coordinates of M is (2, 3, –1)

Now, $P M = \sqrt{{(2 - 0)}^{2} + {(3 - 2)}^{2} + {(- 1 - 3)}^{2}} = \sqrt{4 + 1 + 16} = \sqrt{21}$

$∴$ Area of $△ P Q R = \frac{1}{2} \times Q R \times P M = \frac{1}{2} \times 5 \times \sqrt{21} = \frac{m}{n}$

$\Rightarrow 2 m - 5 \sqrt{21} n = 0$ .

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