If lines x=y=z and x=y/2=z/3 and third line passing through (1,1,1) form a triangle of area 6 units, then point of intersection of third line with second line will be

A
$(1, 2, 3)$
B
$(2, 4, 6)$
C
$(\frac{4}{3}, \frac{8}{3}, \frac{12}{3})$
D
$(2, 1, 3)$

Solution:

x = y = z--- (1), $x = \frac{y}{2} = \frac{z}{3}$ --- (2)
Clearly point of intersection of (1) and (2) is (0,0,0)
D.r’s of (1) are (1, 1, 1)
D.r’s of (2) are (1, 2, 3)
Let $θ$ be the angle between (1) and (2)
$\cos θ = \frac{6}{\sqrt{42}}, \sin θ = \frac{\sqrt{6}}{\sqrt{42}}$
Let any point on second line be $(λ, 2 λ, 3 λ)$
Third line passing through (1, 1, 1)
(1, 1, 1) lies on (1)
A = (1, 1, 1)

$A r e a o f Δ O A B = \frac{1}{2} (O A) O B \sin θ$

$= \frac{1}{2} \sqrt{3} λ \sqrt{14} \times \frac{\sqrt{6}}{\sqrt{42}} = \sqrt{6}$ $\Rightarrow λ = 2$

So B is $(2, 4, 6)$

If lines $x = y = z$ and $x = y / 2 = z / 3$ and third line passing through $(1, 1, 1)$ form a triangle of area $\sqrt{6}$ units, then point of intersection of third line with second line will be

Solution:

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