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 $2 \mathrm{a}+\mathrm{b}+\mathrm{c}-7$ is
Solution:
Required plane is $\mathrm{p}_{1}+\lambda \mathrm{p}_{2}=(2+3 \lambda) \mathrm{x}-(7+5 \lambda) \mathrm{y}$
$+(4+4 \lambda) \mathrm{z}-3+11 \lambda=0$
which is satisfied by $(-2,1,3)$.
Hence, $\lambda=\frac{1}{6}$
Thus, plane is $15 x-47 y+28 z-7=0$
So, $2 a+b+c-7=4$