A line L passing through the point P (1, 4, 3), is perpendicular to both the lines $\frac{x-1}{2}=\frac{y+3}{1}=\frac{z-2}{4},$and $\frac{x+2}{3}=\frac{y-4}{2}=\frac{z+1}{-2}$. If the position vector of point Q on L is$(a_1,a_2,a_3)$ such that ${\left(PQ\right)}^2=357,$then $(a_1+a_2+a_3)$can be

Equation of the line passing through P (1, 4, 3) is $\frac{x-1}{a}=\frac{y-4}{b}=\frac{z-3}{c}\mathrm{.......}\left(1\right)$

Since (1) is perpendicular to $\frac{x-1}{2}=\frac{y+3}{1}=\frac{z-2}{4}and\frac{x+2}{3}=\frac{y-4}{2}=\frac{z+1}{-2}$

Hence $2a+b+4c=0and3a+2b-2c=0$

$\frac{a}{-2-8}=\frac{b}{12+4}=\frac{c}{4-3}$

$\Rightarrow \frac{a}{-10}=\frac{b}{16}=\frac{c}{1}$

Hence the equation of the lines is $\frac{x-1}{-10}=\frac{y-4}{16}=\frac{z-3}{1}\mathrm{....}\left(2\right)$

Ans, Now any point Q on (2) can be taken as $(1-10t,16t+4,t+3)$

Distance of Q from P (1, 4, 3)

$={\left(10t\right)}^2+{\left(16t\right)}^2+t^2=357$

$\Rightarrow (100+256+1)1^2=357\Rightarrow 1=1or-1$

Q is $(-9,20,4)or(11,-12,2)$

Hence $a_1+a_2+a_3=15or1$