The triangle whose vertices are A3,−1,2,Bλ,−1,1,C1,1,−2, is a right-angled triangle. Then the sum of all possible integral values of λ is

# The triangle whose vertices are $A\left(3,-1,2\right),B\left(\lambda ,-1,1\right),C\left(1,1,-2\right)$, is a right-angled triangle. Then the sum of all possible integral values of $\lambda$ is

1. A

$3$

2. B

$-3$

3. C

$-2$

4. D

$2$

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### Solution:

Direction ratios of  $\overline{AB},\overline{BC}\text{\hspace{0.17em}},\overline{CA}$are respectively are $〈3-\lambda ,0,1〉,〈\lambda -1,-2,3〉,〈-2,2,-4〉$

In right angle triangle, right angle may occur at any $B$vertex   $\overline{AB},\overline{BC}\text{\hspace{0.17em}}$

If the right angle is at the vertex then the dot product of  is zero.

$\begin{array}{c}\left(3-\lambda \right)\left(\lambda -1\right)+3=0\\ 3\lambda -3-{\lambda }^{2}+\lambda +3=0\\ {\lambda }^{2}-4\lambda =0\\ \lambda \left(\lambda -4\right)=0\end{array}$

It gives$\lambda =0,4$

If the right angle is at the vertex $C$, then the dot product of   $\overline{AC},\overline{BC}\text{\hspace{0.17em}}$ is zero

$\begin{array}{c}-2\left(3-\lambda \right)-4-12=0\\ 2\lambda -2+16=0\\ 2\lambda +14=0\\ \lambda =-7\end{array}$

It gives  $\lambda =-7$

If the right angle is at the vertex $A$then the dot product of   $\overline{AC},\overline{AB}\text{\hspace{0.17em}}$ is zero

$\begin{array}{c}-2\left(3-\lambda \right)+2\left(0\right)-4\left(1\right)=0\\ -6+2\lambda -4=0\\ 2\lambda =10\\ \lambda =5\end{array}$

It gives $\lambda =5$

Therefore, sum of all the possible values of  $\lambda$ is   $\overline{)2}$

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