If planes x+4y−2z=1, x+7y−5z=β and x+5y+αz=5 intersect in a line in R3 then α+β is equal to

# If planes  and $x+5y+\alpha z=5$ intersect in a line in ${R}^{3}$ then $\alpha +\beta$ is equal to

1. A
2. B
3. C
4. D

Register to Get Free Mock Test and Study Material

+91

Verify OTP Code (required)

I agree to the terms and conditions and privacy policy.

### Solution:

Given planes intersect in a line. This means that

the system of equations  and

$x+5y+\alpha z=5$ has infinitely many solutions.

Now, $D=0⇒\left|\begin{array}{ccc}1& 4& -2\\ 1& 7& -5\\ 1& 5& \alpha \end{array}\right|=0$

$\begin{array}{l}⇒ \left(7\alpha +25\right)-4\left(\alpha +5\right)-2\left(5-7\right)=0⇒3\alpha +9=0⇒\alpha =-3\\ {D}_{3}=0⇒\left|\begin{array}{lll}1& 4& 1\\ 1& 7& \beta \\ 1& 5& 5\end{array}\right|=0\\ ⇒ \left(35-5\beta \right)-4\left(5-\beta \right)+2\left(5-7\right)=0⇒\beta =13\\ \therefore \alpha +\beta =-3+13=10\end{array}$

Register to Get Free Mock Test and Study Material

+91

Verify OTP Code (required)

I agree to the terms and conditions and privacy policy.