Let A be the point of intersection of the lines and . Let B and C be the points on the lines and respectively such that . Then the square of the area of the triangle ABC is [2025]

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Solution

Q.
Let A be the point of intersection of the lines $L_{1} : \frac{x - 7}{1} = \frac{y - 5}{0} = \frac{z - 3}{- 1}$ and $L_{2} : \frac{x - 1}{3} = \frac{y + 3}{4} = \frac{z + 7}{5}$ . Let B and C be the points on the lines $L_{1}$ and $L_{2}$ respectively such that $A B = A C = \sqrt{15}$ . Then the square of the area of the triangle ABC is [2025]

1 57

2 54

3 63

4 60

Ans.
(2)

We have, $L_{1} : \frac{x - 7}{1} = \frac{y - 5}{0} = \frac{z - 3}{- 1}$ and $L_{2} : \frac{x - 1}{3} = \frac{y + 3}{4} = \frac{z + 7}{5}$

Angle between $L_{1}$ and $L_{2}$ ,

$\cos θ =$ $| \frac{3 \times 1 + 4 \times 0 + 5 (- 1)}{\sqrt{1 + 0 + 1} \sqrt{9 + 16 + 25}} |$ $= \frac{1}{5}$

$\Rightarrow \sin θ = \frac{\sqrt{24}}{5}$

Now, area of $△ A B C = \frac{1}{2} a b \sin θ = \frac{1}{2} {(\sqrt{15})}^{2} (\frac{\sqrt{24}}{5})$

So, square of area = $\frac{15 \times 15 \times 24}{4 \times 25} = 54$ .

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