Find the equation of the plane which contains the line of intersection of the planes $\vec r \cdot (\hat i + 2\hat j + 3\hat k) - 4 = 0,\vec r \cdot (2\hat i + \hat j - \hat k) + 5 = 0$ and which is perpendicular to the plane $\vec r \cdot (5\hat i + 3\hat j - 6\hat k) + 8 = 0.$

.: The given planes are
$\vec r \cdot (\hat i + 2\hat j + 3\hat k) - 4 = 0,\vec r \cdot (2\hat i + \hat j - \hat k) + 5 = 0$and

$\vec r \cdot (5\hat i + 3\hat j - 6\hat k) + 8 = 0.$

Putting $\vec r = x\hat i + y\hat j + z\hat k,$

we get
$x + 2y + 3z - 4 = 0$ ..(1)

$2x + y - z + 5 = 0$ ...(2)

and $5x + 3y - 6z + 8 = 0$ ...(3)

Any plane passing through the intersection of (1) and (2) is

$(x + 2y + 3z - 4) + k(2x + y - z + 5) = 0$

$\Rightarrow$ $(1 + 2k)x + (2 + k)y + (3 - k)z + ( - 4 + 5k) = 0$ ...(4)

Now, plane (4) is perpendicular to plane (3),

$\therefore$ $5(1 + 2k) + 3(2 + k) - 6(3 - k) = 0$

$\Rightarrow$ $5 + 10k + 6 + 3k - 18 + 6k = 0$

$\Rightarrow$ $19k - 7 = 0 \Rightarrow k = 7/19$

Putting in (4),

we get $33x + 45y + 50z - 41 = 0,$which is the required equation of the plane.