A line d.c's proportional to $(2, 1, 2)$ meets each of the lines $x = y + a = z$ and $x + a = 2 y = 2 z$ . Then the coordinates of each of the points of intersection are given by

(3 a, 2 a, 3 a); (a, a, 2 a)

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(3 a, 2 a, 3 a); (a, a, a)

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(3 a, 3 a, 3 a); (a, a, a)

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(2 a, 3 a, 3 a); (2 a, a, a)

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Solution

The correct option is B $(3 a, 2 a, 3 a); (a, a, a)$
$L_{1} : \frac{x}{1} = \frac{y + a}{1} = \frac{z}{1}$
$L_{2} : \frac{x + a}{2} = \frac{y}{1} = \frac{z}{1}$
Any point on $L_{1} A (k_{1}, k_{1} - 1, k_{1})$
Any point on $L_{2} B (2 k_{2} - 1, k_{2}, k_{2})$
Drs of $A B = (k_{1} - 2 k_{2} + a, k_{1} - k_{2} - a, k_{1} - k - 2)$
$= \frac{k_{1} - 2 k_{2} + a}{2} = \frac{k_{1} - k_{2} - a}{1} = \frac{k_{1} - k_{2}}{2}$
$\Rightarrow k_{1} - 2 k_{2} + a = 2 k_{1} - 2 k_{2} - 2 a$
$\Rightarrow 3 a = k_{1}$
$∴ 2 k_{1} - 2 k_{2} - 2 a = k_{1} - k_{2}$
$\Rightarrow 2 k_{1} - k_{1} - k_{2} = 2 a$
$\Rightarrow k_{1} - k_{2} = 2 a$
$\Rightarrow k_{2} = a$
$∴ A (3 a, 2 a, 3 a)$
$B (a, a, a)$

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