Let d be the distance of the point of interse | vedclass

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(C) Let the first line be $\frac{x+6}{3}=\frac{y}{2}=\frac{z+1}{1}=\lambda$. The general point on this line is $(3\lambda-6, 2\lambda, \lambda-1)$.Let

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$75$

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(C) Let the first line be $\frac{x+6}{3}=\frac{y}{2}=\frac{z+1}{1}=\lambda$. The general point on this line is $(3\lambda-6, 2\lambda, \lambda-1)$.Let the second line be $\frac{x-7}{4}=\frac{y-9}{3}=\frac{z-4}{2}=\mu$. The general point on this line is $(4\mu+7, 3\mu+9, 2\mu+4)$.Equating the coordinates for the point of intersection:$3\lambda-6 = 4\mu+7 \Rightarrow 3\lambda-4\mu = 13$ (Equation $1$)$2\lambda = 3\mu+9 \Rightarrow 2\lambda-3\mu = 9$ (Equation $2$)Solving these equations:Multiply Equation $1$ by $2$ and Equation $2$ by $3$:$6\lambda-8\mu = 26$$6\lambda-9\mu = 27$Subtracting the two equations gives $\mu = -1$.Substituting $\mu = -1$ into Equation $2$: $2\lambda - 3(-1) = 9 \Rightarrow 2\lambda = 6 \Rightarrow \lambda = 3$.The point of intersection is $(3(3)-6, 2(3), 3-1) = (3, 6, 2)$.The distance $d$ from $(3, 6, 2)$ to $(7, 8, 9)$ is given by:$d^2 = (7-3)^2 + (8-6)^2 + (9-2)^2 = 4^2 + 2^2 + 7^2 = 16 + 4 + 49 = 69$.Therefore,$d^2+6 = 69+6 = 75$.

Let $d$ be the distance of the point of intersection of the lines $\frac{x+6}{3}=\frac{y}{2}=\frac{z+1}{1}$ and $\frac{x-7}{4}=\frac{y-9}{3}=\frac{z-4}{2}$ from the point $(7,8,9)$. Then $d^2+6$ is equal to :

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