Question

If the length of the perpendicular drawn from the point P(a, 4, 2), a> 0 on the line
$\frac{x+1}{2} = \frac{y-3}{3} = \frac{z-1}{1}$ is $2\sqrt6$ units and $Q(α1, α2, α3)$
is the image of the point P in this line, then
$\alpha + \sum_{i=1}^{3} \alpha_i$
is equal to :

• A. 7
• B. 8
• C. 12
• D. 14

Answer 1

Answer: (B)

8

Answer 2

The correct answer is (B) : 8
∵ PR is perpendicular to given line, so

Fig.

$2(2λ-1-α)+3(3λ-1)-1(-λ-1)=0$
$⇒ α = 7λ-2$
Now,
$∵ PR = 2\sqrt6$
$⇒ (-5λ+1)^2 + (3λ-1)^2 + (λ+1)^2 = 24$
$⇒ 5λ^2 - 2λ-3 = 0$
$⇒ λ = 1\ or -\frac{3}{5}$
$∵ α>0\ so λ = 1\ and α = 5$
Now $\sum_{i=1}^{3} \alpha_i$ = 2(Sum of co-ordinate of R)-(Sum of co-ordinates of P)
= 2(7)-11
= 3
a+$\sum_{i=1}^{3} \alpha_i$= 5+3
= 8