If the length of the perpendicular drawn from the point P(a,4,2),a>0 on the line x+1 2= y-3 3= z-1 -1 is 2√6 units and Q(α _1,α _2,α _3) is the image…

Let a point on the line $\frac{x+1}{2}=\frac{y-3}{3}=\frac{z-1}{-1}=\lambda$2x+1​=3y−3​=−1z−1​=λ be

$$M(2\lambda -1,3\lambda +3,-\lambda +1)$$M(2λ−1,3λ+3,−λ+1)

and the DR's of the line are $<2,3,-1>$<2,3,−1>

The DR's of line joining the points $(2\lambda -1,3\lambda +3,-\lambda +1)$(2λ−1,3λ+3,−λ+1) and $(a,4,2)$(a,4,2) will be

$$<2\lambda -1-a,3\lambda -1,-\lambda -1>$$<2λ−1−a,3λ−1,−λ−1>

Now its perpendicular to the given line, so$(2\lambda -1-a)2+(3\lambda -1)3+(-\lambda -1)(-1)=0$(2λ−1−a)2+(3λ−1)3+(−λ−1)(−1)=0

$$\Rightarrow 4\lambda -2-2a+9\lambda -3+\lambda +1=0$$⇒4λ−2−2a+9λ−3+λ+1=0

$$\Rightarrow 14\lambda -4-2a=0$$⇒14λ−4−2a=0

$$\Rightarrow 7\lambda -2-a=0\Rightarrow 7\lambda -2=a$$⇒7λ−2−a=0⇒7λ−2=a

and,

$${(2\lambda -1-a)}^{2}+{(3\lambda -1)}^{2}+{(\lambda +1)}^{2}={(2\sqrt{6})}^{2}$$(2λ−1−a)2+(3λ−1)2+(λ+1)2=(26​)2

$$\Rightarrow {(5\lambda -1)}^{2}+{(3\lambda -1)}^{2}+{(\lambda +1)}^{2}=24$$⇒(5λ−1)2+(3λ−1)2+(λ+1)2=24

$$\Rightarrow 35{\lambda }^{2}-14\lambda -21=0$$⇒35λ2−14λ−21=0

$$\Rightarrow (\lambda -1)(35\lambda +21)=0$$⇒(λ−1)(35λ+21)=0

For, $\lambda =1;a=5$λ=1;a=5

i.e. $M\equiv (1,6,0)$M≡(1,6,0)

As $Q({\alpha }_{1},{\alpha }_{2},{\alpha }_{3})$Q(α1​,α2​,α3​) is the image of point $P$P

$$\begin{matrix}\frac{{\alpha }_{1}+5}{2}=1; & \frac{{\alpha }_{2}+4}{2}=6; & \frac{{\alpha }_{3}+2}{2}=0\end{matrix}$$2α1​+5​=1;​2α2​+4​=6;​2α3​+2​=0​

$$\Rightarrow \begin{matrix}{\alpha }_{1}=-3; & {\alpha }_{2}=8; & {\alpha }_{3}=-2\end{matrix}$$⇒α1​=−3;​α2​=8;​α3​=−2​

Hence, $a+{\alpha }_{1}+{\alpha }_{2}+{\alpha }_{3}=8$a+α1​+α2​+α3​=8