Q.Find the vector equation of the plane passing through the intersection of the planes vector r . (2i + 2j - 3k) = 7 and vector r . (2i + 5j + 3k) = 9 and the point (2, 1, 3). OR Find the shortest distance between the straight lines (x-3)/3 = (y-8)/-1 = (z-3)/1 and (x+3)/-3 = (y+7)/2 = (z-6)/4.

Write the family of planes through the intersection of the two given planes with parameter $\lambda$, plug in the given point to solve for $\lambda$, then substitute back.

Given planes: $\vec r\cdot(2\hat i+2\hat j-3\hat k)=7$ and $\vec r\cdot(2\hat i+5\hat j+3\hat k)=9$, and the plane must pass through the point $(2,1,3)$.

Step 1 — family of planes through the intersection:

$\left[\vec r\cdot(2\hat i+2\hat j-3\hat k) - 7\right] + \lambda\left[\vec r\cdot(2\hat i+5\hat j+3\hat k)-9\right] = 0$

$\vec r\cdot\big[(2+2\lambda)\hat i + (2+5\lambda)\hat j + (-3+3\lambda)\hat k\big] = 7+9\lambda$

Step 2 — impose the point $(2,1,3)$, i.e. $\vec r = 2\hat i+\hat j+3\hat k$:

$2(2+2\lambda) + 1(2+5\lambda) + 3(-3+3\lambda) = 7+9\lambda$

$(4+4\lambda) + (2+5\lambda) + (-9+9\lambda) = 7+9\lambda$

$-3 + 18\lambda = 7+9\lambda \Rightarrow 9\lambda = 10 \Rightarrow \lambda = \dfrac{10}{9}$

Step 3 — substitute back:

Coefficient of $\hat i$: $2+2\lambda = 2+\frac{20}{9}=\frac{38}{9}$

…