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Question

(A) Let the given line be $\frac{x+2}{1}=\frac{y-1}{2}=\frac{z+1}{-2}=\lambda$. Any point on this line is given by $(\lambda-2, 2\lambda+1, -2\lambda-

Vedclass · Accepted Answer

$(2, 9, -9), (-6, -7, 7)$

Vedclass · Answer

(A) Let the given line be $\frac{x+2}{1}=\frac{y-1}{2}=\frac{z+1}{-2}=\lambda$. Any point on this line is given by $(\lambda-2, 2\lambda+1, -2\lambda-1)$. The distance of this point from $A(-2, 1, -1)$ is $12$. Using the distance formula: $\sqrt{(\lambda-2 - (-2))^2 + (2\lambda+1 - 1)^2 + (-2\lambda-1 - (-1))^2} = 12$. $\sqrt{\lambda^2 + (2\lambda)^2 + (-2\lambda)^2} = 12$. $\sqrt{\lambda^2 + 4\lambda^2 + 4\lambda^2} = 12$. $\sqrt{9\lambda^2} = 12$. $3|\lambda| = 12$,so $\lambda = \pm 4$. For $\lambda = 4$,the point is $(4-2, 2(4)+1, -2(4)-1) = (2, 9, -9)$. For $\lambda = -4$,the point is $(-4-2, 2(-4)+1, -2(-4)-1) = (-6, -7, 7)$. Thus,the required points are $(2, 9, -9)$ and $(-6, -7, 7)$.

The coordinates of the points on the line $\frac{x+2}{1}=\frac{y-1}{2}=\frac{z+1}{-2}$ at a distance of $12 \text{ units}$ from the point $A(-2, 1, -1)$ are

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