Let the line x 1= 6-y 2= z+8 5 intersect the lines x-5 4= y-7 3= z+2 1 and x+3 6= 3-y 3= z-6 1 at the points A and B respectively. Then the distance…

Given,

The line $\frac{x}{1}=\frac{6-y}{2}=\frac{z+8}{5}$1x​=26−y​=5z+8​ intersect the lines $\frac{x-5}{4}=\frac{y-7}{3}=\frac{z+2}{1}$4x−5​=3y−7​=1z+2​ and $\frac{x+3}{6}=\frac{3-y}{3}=\frac{z-6}{1}$6x+3​=33−y​=1z−6​ at the points $A$A and $B$B respectively,

Now plotting the diagram, we get,

Now solving, $\frac{x}{1}=\frac{y-6}{-2}=\frac{z+8}{5}=\lambda$1x​=−2y−6​=5z+8​=λ and $\frac{x-5}{4}=\frac{y-7}{3}=\frac{z+2}{1}=t$4x−5​=3y−7​=1z+2​=t we get,

$$A\equiv (\lambda ,-2\lambda +6,5\lambda -8)\equiv (4t+5,3t+7,t-2)$$A≡(λ,−2λ+6,5λ−8)≡(4t+5,3t+7,t−2)

Now comparing both side and solving we get, $t=-1,\lambda =1$t=−1,λ=1

Hence, point$A(1,4,–3)$A(1,4,–3)

Now solving, $\frac{x}{1}=\frac{y-6}{-2}=\frac{z+8}{5}=\mu$1x​=−2y−6​=5z+8​=μ and $\frac{x+3}{6}=\frac{y-3}{-3}=\frac{z-6}{1}=k$6x+3​=−3y−3​=1z−6​=k we get,

For point $B\equiv (\mu ,-2\mu +6,5\mu -8)$B≡(μ,−2μ+6,5μ−8) $\equiv (6k-3,-3k+3,k+6)$≡(6k−3,−3k+3,k+6)

Now comparing both side and solving we get, $k=1,\mu =3$k=1,μ=3

Hence, point $B(3,0,7)$B(3,0,7)

So, mid-point of $AB=M(2,2,2)$AB=M(2,2,2)

Now finding, distance of $M$M from the plane $2x–2y+z–14=0$2x–2y+z–14=0 we get,

$$|\frac{4-4+2-14}{\sqrt{{2}^{2}+{2}^{2}+{1}^{2}}}|=|\frac{-12}{3}|=4$$∣22+22+12​4−4+2−14​∣=∣3−12​∣=4