Let $L_{1} : x = y = z$
$L_{2} : x - 1 = y - 2 = z - 3$ be two lines
Let from origin O(0, 0, 0) on $L_{1},$ perpendicular is drawn to $L_{2}$ has foot ‘A’. Segment OA is rotated about O by an angle $90^{0}$ such that $L_{2}$ moves along with it, without changing its direction and becomes $L_{3} .$ ‘A’ becomes $B (α, β, ϒ)$ then $α + β + y =$

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Detailed Solution

Let $A (λ + 1, λ + 2, λ + 3)$

$∴ (λ + 1) + (λ + 2) + (λ + 3) = 0 \Rightarrow λ = - 6 \Rightarrow A (- 5, - 4, - 3)$

Let line $L_{1}$ and $L_{2}$ lies on plane P. Then normal to the plane P will be 5i + 4j + 3k.

$\Rightarrow$ Equation of plane P is 5x + 4y + 3z = 0

$∴$ Dr’s of line through origin and $⊥$ to $L_{3}$ is $(5 i + 4 j + 3 k) (i + j + k) = < 1, - 2, 1 >$

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