Question

Let the equation of a line is $\frac{x-2}{1}=\frac{y-3}{2}=\frac{z-4}{3}$ . An insect starts flying from $P\left(1,3,2\right)$ in a straight line meeting the given line at a point $R\left(a,b,c\right)$ and then goes to the point $Q\left(6,7,5\right)$ in a straight line such that $PR$ is perpendicular to $RQ.$ Then, the least value of $7\left(a+b+c\right)$ is equal to

• A. $105$
• B. $45$
• C. $10$
• D. $7$

Answer 1

Answer: (B)

Any general point on the line is $\left(\lambda +2,2\lambda +3,\lambda +4\right)$

$\overrightarrow{PR}=<\lambda +1,2\lambda ,3\lambda +2>$
$\overrightarrow{RQ}=<4-\lambda ,4-2\lambda ,1-3\lambda >$
From the diagram, $\overrightarrow{PR}\cdot \overrightarrow{RQ}=0$
$\Rightarrow \left(\lambda +1\right)\left(4-\lambda \right)+2\lambda \left(4-2\lambda \right)+\left(3\lambda +2\right)\left(1-3\lambda \right)=0$
$\Rightarrow {\lambda }^2-3\lambda -4+4{\lambda }^2-8\lambda +9{\lambda }^2+3\lambda -2=0$
$\Rightarrow 14{\lambda }^2-8\lambda -6=0$
$\Rightarrow \lambda =1,-\frac{3}{7}$
Coordinates of $R$ are $\left(3,5,7\right)$ or $\left(\frac{11}{7},\frac{15}{7},\frac{19}{7}\right)$
Least value of $7\left(a+b+c\right)=11+15+19=45$