Three Dimensional Geometry — Class 12 Maths Solution

.: The given planes are $\vec r \cdot (2\hat i + 2\hat j - 3\hat k) = 7$and $\vec r \cdot (2\hat i + 5\hat j + 3\hat k) = 9$,
Putting $\vec r = x\hat i + y\hat j + z\hat k,$

we get
$(x\hat i + y\hat j + z\hat k) \cdot (2\hat i + 2\hat j - 3\hat k) = 7$

and
$(x\hat i + y\hat j + z\hat k) \cdot (2\hat i + 5\hat j + 3\hat k) = 9$

i.e. $2x + 2y - 3z - 7 = 0$ ...(1)
and $2x + 5y + 3z - 9 = 0$ ...(2)
Any plane through the intersection of (1) and (2) is
$(2x + 2y - 3z - 7) + k(2x + 5y + 3z - 9) = 0$ ...(3)
Since it passes through (2, 1, 3),
$\therefore$ $(4 + 2 - 9 - 7) + k(4 + 5 + 9 - 9 = 0$

$\Rightarrow$ $- 10 + k(9) = 0 \Rightarrow k = \cfrac{{10}}{9}$ .
Putting in (3),

we get
$(2x + 2y - 3z - 7) + \cfrac{{10}}{9}(2x + 5y + 3z - 9) = 0$

$\Rightarrow$ $18x + 18y - 27z - 63 + 20x + 50y + 30z - 90 = 0$

$\Rightarrow$ $38x + 68y + 3z - 153 = 0$
$\Rightarrow$ $(x\hat i + y\hat j + z\hat k) \cdot (38\hat i + 68\hat j + 3\hat k) = 153$

$\Rightarrow$ $\vec r \cdot (38\hat i + 68\hat j + 3\hat k) = 153$, which is the required equation.