If the equation of the plane

Question:

If the equation of the plane passing through the line of intersection of the planes $2 x-7 y+4 z-3=0,3 x-5 y+4 z+11=0$ and the point $(-2,1,3)$ is $a x+b y+c z-7=0$, then the value of $2 a+b+c-7$ is

Solution:

Required plane is

$\mathrm{p}_{1}+\lambda \mathrm{p}_{2}=(2+3 \lambda) \mathrm{x}-(7+5 \lambda) \mathrm{y}$

$+(4+4 \lambda) z-3+11 \lambda=0$

which is satisfied by $(-2,1,3)$

Hence, $\lambda=\frac{1}{6}$

Thus, plane is $15 x-47 y+28 z-7=0$

So, $2 a+b+c-7=4$

Leave a comment

Close
faculty

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now