Let PQR be a triangle with R(–1, 4, 2). Suppose M(2, 1, 2) is the mid point of PQ. The distance of the centroid of from the point of intersection of the lines and is [2024]

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Solution

Q.
Let PQR be a triangle with R(–1, 4, 2). Suppose M(2, 1, 2) is the mid point of PQ. The distance of the centroid of $△ P Q R$ from the point of intersection of the lines $\frac{x - 2}{0} = \frac{y}{2} = \frac{z + 3}{- 1}$ and $\frac{x - 1}{1} = \frac{y + 3}{- 3} = \frac{z + 1}{1}$ is [2024]

1 $\sqrt{69}$

2 69

3 $\sqrt{99}$

4 9

Ans.
(1)

Let centroid G divides MR in the ratio 1 : 2.

So, centroid is given by

$∴ G (\frac{4 - 1}{3}, \frac{2 + 4}{3}, \frac{4 + 2}{3})$ i.e., G(1, 2, 2)

Let $l_{1} : \frac{x - 2}{0} = \frac{y}{2} = \frac{z + 3}{- 1} = λ$ (say)

$\Rightarrow x = 2, y = 2 λ, z = - λ - 3$

$l_{2} : \frac{x - 1}{1} = \frac{y + 3}{- 3} = \frac{z + 1}{1} = r$ (say)

$\Rightarrow$ x = r + 1, y = –3r – 3, z = r – 1

Now, $2 = r + 1 \Rightarrow r = 1$ and $2 λ = - 3 r - 3 \Rightarrow λ = - 3$ .

$∴$ Point of intersection of lines $l_{1}$ and $l_{2}$ is given by A(2, –6, 0).

$∴$ Required distance, $A G = \sqrt{1 + 64 + 4} = \sqrt{69}$ .

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