Question

If the length of the perpendicular drawn from the point $P(a,4,2),a>0$ on the line $\frac{x+1}{2}=\frac{y-3}{3}=\frac{z-1}{-1}\text{is}2\sqrt{6}$ units and $Q({\alpha }_1,{\alpha }_2,{\alpha }_3)$ is the image of the point $P$ in this line, then $a+\sum _{i=1}^3{\alpha }_i$ is equal to :

• A. $14$
• B. $8$
• C. $12$
• D. $7$

Answer 1

The correct option is B $8$

$\because PR$ is perpendicular to given line, so

$2(2\lambda -1-a)+3(3\lambda -1)-1(-\lambda -1)=0$
$\Rightarrow a=7\lambda -2$
Now
$\because PR=2\sqrt{6}$
$\Rightarrow (-5\lambda +1{)}^2+(3\lambda -1{)}^2+(\lambda +1{)}^2=24$
$\Rightarrow 5{\lambda }^2-2\lambda -3=0\Rightarrow \lambda =1\text{or}-\frac{3}{5}$
$\because a>0\text{so}\lambda =1\text{and}a=5$
Now $\sum _{i=1}^3{\alpha }_i=2\text{(Sum of co-ordinate of}R)-\text{(Sum of co-ordinate of}P)$
$=2\left(7\right)-11=3$
$a+\sum _{i=1}^3{\alpha }_i=5+3=8$