Let P and Q be the points on the line which are at a distance of 6 units from the point R(1, 2, 3). If the centroid of the triangle PQR is , then is : [2024]

Question

Let P and Q be the points on the line $\frac{x + 3}{8} = \frac{y - 4}{2} = \frac{z + 1}{2}$ which are at a distance of 6 units from the point R(1, 2, 3). If the centroid of the triangle PQR is $(α, β, γ)$ , then $α^{2} + β^{2} + γ^{2}$ is : [2024]

Smart Achievers · Accepted Answer

(1)

The given equation of line is,

$\frac{x + 3}{8} = \frac{y - 4}{2} = \frac{z + 1}{2}$ (say)

$(α, β, γ)$ Any point on line is $(8 λ - 3, 2 λ + 4, 2 λ - 1)$

Now, distance of this point from R(1, 2, 3)

$\Rightarrow {(8 λ - 4)}^{2} + {(2 λ + 2)}^{2} + {(2 λ - 4)}^{2} = 36$

$\Rightarrow 72 λ^{2} - 72 λ = 0 \Rightarrow λ (λ - 1) = 0$

$\Rightarrow λ = 0 or λ = 1$

So, coordinates of P = (–3, 4, –1) and coordinates of Q = (5, 6, 1).

Now, centroid of $△ P Q R$ = (1, 4, 1) $\Rightarrow α = 1, β = 4, γ = 1$

So, $α^{2} + β^{2} + γ^{2} = 18$ .

Solution

Q.
Let P and Q be the points on the line $\frac{x + 3}{8} = \frac{y - 4}{2} = \frac{z + 1}{2}$ which are at a distance of 6 units from the point R(1, 2, 3). If the centroid of the triangle PQR is $(α, β, γ)$ , then $α^{2} + β^{2} + γ^{2}$ is : [2024]

1 18

2 24

3 26

4 36

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