Q.(a) Find the equation of the plane through the intersection of the planes 3x − y + 2z − 4 = 0 and x + y + z − 2 = 0 and the point (2, 2, 1). (2 marks)

The family of planes through the intersection of two given planes is $(\text{P}_1)+\lambda(\text{P}_2)=0$; substituting the given point fixes $\lambda$. The two Cartesian lines are read off directly as vector equations, and the shortest distance between skew lines uses the standard triple-product/cross-product formula.

(a) Family of planes through the intersection of $3x-y+2z-4=0$ and $x+y+z-2=0$:

$(3x-y+2z-4) + \lambda(x+y+z-2) = 0$

Substitute the point $(2,2,1)$:

$(3(2)-2+2(1)-4) + \lambda(2+2+1-2) = 0$

$(6-2+2-4) + \lambda(3) = 0$

$2+3\lambda=0 \Rightarrow \lambda=-\dfrac23$

Plane: $(3x-y+2z-4) - \dfrac23(x+y+z-2)=0$. Multiply by 3:

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

$9x-3y+6z-12-2x-2y-2z+4=0$

$7x-5y+4z-8=0$

(Check: at $(2,2,1)$, $14-10+4-8=0$ ✓.)

(b) Line 1: $\dfrac{x+1}{7}=\dfrac{y+1}{-6}=\dfrac{z+1}{1}$ passes through $(-1,-1,-1)$ with direction ratios $(7,-6,1)$:

$\vec r = (-\hat i-\hat j-\hat k) + s(7\hat i-6\hat j+\hat k)$

Line 2: $\dfrac{x-3}{1}=\dfrac{y-5}{-2}=\dfrac{z-7}{1}$ passes through $(3,5,7)$ with direction ratios $(1,-2,1)$:

$\vec r = (3\hat i+5\hat j+7\hat k) + t(\hat i-2\hat j+\hat k)$

(c) For skew lines $\vec r=\vec b_1+s\vec d_1$ and $\vec r=\vec b_2+t\vec d_2$, shortest distance $= \dfrac{|(\vec b_2-\vec b_1)\cdot(\vec d_1\times\vec d_2)|}{|\vec d_1\times\vec d_2|}$.

$\vec b_2-\vec b_1 = (3-(-1),5-(-1),7-(-1)) = (4,6,8)$

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