Prove the following

Question:

If for some $\alpha$ and $\beta$ in $\mathbf{R}$, the intersection of the following three planes

$x+4 y-2 z=1$

$x+7 y-5 z=\beta$

$x+5 y+\alpha z=5$

is a line in $R^{3}$, then $\alpha+\beta$ is equal to:

 

  1. 0

  2. 10

  3. 2

  4. $-10$


Correct Option: , 2

Solution:

$\Delta=0 \Rightarrow\left|\begin{array}{ccc}1 & 4 & -2 \\ 1 & 7 & -5 \\ 1 & 5 & \alpha\end{array}\right|=0$

$\Rightarrow(7 \alpha+25)-(4 \alpha+10)+(-20+14)=0$

$\Rightarrow 3 \alpha+9=0 \Rightarrow \alpha=-3$

Also, $D_{z}=0 \Rightarrow\left|\begin{array}{lll}1 & 4 & 1 \\ 1 & 7 & \beta \\ 1 & 5 & 5\end{array}\right|=0$

$\Rightarrow 1(35-5 \beta)-(15)+1(4 \beta-7)=0 \Rightarrow \beta=13$

Hence, $\alpha+\beta=-3+13=10$

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