If the lines x - 1 1 = y - 2 2 = z + 3 1 and x - a 2 = y + 2 3 = z - 3 1 intersect at the point P, then the distance of the point P from the plane z = a is [2023]

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Solution

Q.
If the lines $\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z + 3}{1}$ and $\frac{x - a}{2} = \frac{y + 2}{3} = \frac{z - 3}{1}$ intersect at the point P, then the distance of the point P from the plane $z = a$ is [2023]

1 10

2 28

3 22

4 16

Ans.
(2)

$Let two lines L_{1} and L_{2} intersect at point P .$ Then we have the points on line $L_{1}$ are as $= (λ + 1, 2 λ + 2, λ - 3)$ and we have the points on line $L_{2}$ are as $= (2 μ + a, 3 μ - 2, μ + 3)$

${∵ L_{1} = \frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z + 3}{1} = λ (say) and L_{2} = \frac{x - a}{2} = \frac{y + 2}{3} = \frac{z - 3}{1} = μ (say)}$

$Now if L_{1} and L_{2} intersect, then we must have$

$λ - 3 = μ + 3 \Rightarrow λ = μ + 6 ...(i)$

and $2 λ + 2 = 3 μ - 2 \Rightarrow 2 λ = 3 μ - 4 ...(ii)$

On solving (i) and (ii), we get $μ = 16$ and $λ = 22$

So, the coordinates of $P$ must be $(23, 46, 19)$ $\Rightarrow a = - 9$

Hence, the distance of the point $P$ from $z = - 9$ is $28 .$

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