Q.Find the vector equation of the plane passing through the intersection of the plane r.(\u00ee+\u0135+k\u0302) = 6 and r.(2\u00ee+3\u0135+4k\u0302) = -5 and the point (1, 1, 1). OR Find the shortest distance between the lines: (x+1)/7 = (y+1)/-6 = (z+1)/1 and (x-3)/1 = (y-5)/-2 = (z-7)/1

Write the family of planes through the intersection of the two given planes, then use the given point to fix the free parameter.

The family of planes through the line of intersection of $\vec r\cdot(\hat i+\hat j+\hat k)=6$ and $\vec r\cdot(2\hat i+3\hat j+4\hat k)=-5$ is:

$\left[\vec r\cdot(\hat i+\hat j+\hat k)-6\right] + \lambda\left[\vec r\cdot(2\hat i+3\hat j+4\hat k)+5\right] = 0$

which rearranges to:

$\vec r\cdot\big[(1+2\lambda)\hat i+(1+3\lambda)\hat j+(1+4\lambda)\hat k\big] = 6-5\lambda$

The plane passes through the point $(1,1,1)$, i.e. $\vec r=\hat i+\hat j+\hat k$ satisfies this. Substituting:

$(1+2\lambda)+(1+3\lambda)+(1+4\lambda) = 6-5\lambda$

$3+9\lambda = 6-5\lambda \implies 14\lambda = 3 \implies \lambda = \frac{3}{14}$

Substitute $\lambda=\frac{3}{14}$ back into the family equation:

$1+2\lambda = \frac{20}{14},\quad 1+3\lambda=\frac{23}{14},\quad 1+4\lambda=\frac{26}{14},\quad 6-5\lambda = \frac{69}{14}$

Multiplying through by $14$:

$\vec r\cdot(20\hat i+23\hat j+26\hat k) = 69$

OR — shortest distance between two skew lines.

Line 1: point $A=(-1,-1,-1)$, direction $\vec b_1=(7,-6,1)$.

Line 2: point $B=(3,5,7)$, direction $\vec b_2=(1,-2,1)$.

Step 1 — compute $\vec b_1\times\vec b_2$: …