Let the point, on the line passing through the points P(1, –2, 3) and Q(5, –4, 7), farther from the origin and at a distance of 9 units from the point P, be . Then is equal to [2024]

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Solution

Q.
Let the point, on the line passing through the points P(1, –2, 3) and Q(5, –4, 7), farther from the origin and at a distance of 9 units from the point P, be $(α, β, γ)$ . Then $α^{2} + β^{2} + γ^{2}$ is equal to [2024]

1 160

2 165

3 150

4 155

Ans.
(4)

Line through PQ is given by

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

Any point R on PQ will be $R (4 λ + 1, - 2 λ - 2, 4 λ + 3)$

Since, PR = 9 units $\Rightarrow {(P R)}^{2} = 81$

$\Rightarrow {(4 λ + 1 - 1)}^{2} + {(- 2 λ - 2 + 2)}^{2} + {(4 λ + 3 - 3)}^{2} = 81$

$\Rightarrow 16 λ^{2} + 4 λ^{2} + 16 λ^{2} = 81$

$\Rightarrow 36 λ^{2} = 81 \Rightarrow λ^{2} = \frac{9}{4} \Rightarrow λ = \pm \frac{3}{2}$

$∴$ R can be (7, –5, 9) or (–5, 1, –3)

Distance from origin of the points be $\sqrt{49 + 25 + 81}$ and $\sqrt{25 + 1 + 9}$ i.e., $\sqrt{155}, \sqrt{35}$

$∴$ Distance of (7, –5, 9) is farthest from origin

$\Rightarrow$ $(α, β, γ) \equiv (7, - 5, 9)$

Hence $α^{2} + β^{2} + γ^{2} = 7^{2} + {(- 5)}^{2} + 9^{2} = 155$ .

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