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

The Correct Option is B

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