Let L be the line x + 1 2 = y + 1 3 = z + 3 6 and let S be the set of all points (a,b,c) on L, whose distance from the line x + 1 2 = y + 1 3 = z - 9 0 along the line L is 7. Then ( a , b , c ) S ( a + b + c ) is equal to: [2026]

Question

Let L be the line $\frac{x + 1}{2} = \frac{y + 1}{3} = \frac{z + 3}{6}$ and let S be the set of all points (a,b,c) on L, whose distance from the line $\frac{x + 1}{2} = \frac{y + 1}{3} = \frac{z - 9}{0}$ along the line L is 7. Then $\sum_{(a, b, c) \in S} (a + b + c)$ is equal to: [2026]

Smart Achievers · Accepted Answer

(3)

$M is the point of intersection of L_{1} and L_{2}$

$\Rightarrow 2 λ - 1 = 2 μ - 1, 3 λ - 1 = 3 μ - 1, 6 λ - 3 = 9$

$\Rightarrow λ = 2 = μ$

$\Rightarrow M (3, 5, 9)$

$Now let point P be (2 K - 1, 3 K - 1, 6 K - 3) on L_{2},$ $such that P M = 7$

$\Rightarrow \sqrt{{(2 K - 4)}^{2} + {(3 K - 6)}^{2} + {(6 K - 12)}^{2}} = 7$

$\Rightarrow 49 K^{2} + 196 - 196 K = 49$

$\Rightarrow K^{2} + 4 - 4 K = 1$

$\Rightarrow K^{2} - 4 K + 3 = 0$

$\Rightarrow K = 1, 3$

$So points P and Q are (1, 2, 3) and (5, 8, 15)$

$So sum of all coordinates of P and Q = 34$

Solution

Q.
Let L be the line $\frac{x + 1}{2} = \frac{y + 1}{3} = \frac{z + 3}{6}$ and let S be the set of all points (a,b,c) on L, whose distance from the line $\frac{x + 1}{2} = \frac{y + 1}{3} = \frac{z - 9}{0}$ along the line L is 7. Then $\sum_{(a, b, c) \in S} (a + b + c)$ is equal to: [2026]

1 6

2 40

3 34

4 28

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