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(B) The equation of line $L_1$ passing through $(0, -1, 0)$ with direction ratios $(1, 1, 1)$ is given by $\frac{x}{1} = \frac{y+1}{1} = \frac{z}{1} =

Vedclass · Accepted Answer

$8$

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(B) The equation of line $L_1$ passing through $(0, -1, 0)$ with direction ratios $(1, 1, 1)$ is given by $\frac{x}{1} = \frac{y+1}{1} = \frac{z}{1} = \lambda$. Thus,the coordinates of point $A$ are $(\lambda, \lambda-1, \lambda)$. The line segment $AB$ connects $A(\lambda, \lambda-1, \lambda)$ and $B(1, 1, 1)$. The direction ratios of line $AB$ are $(\lambda-1, \lambda-2, \lambda-1)$. Given that the direction ratios of this line are proportional to $2, 1, 2$,we have: $\frac{\lambda-1}{2} = \frac{\lambda-2}{1} = \frac{\lambda-1}{2}$. From $\frac{\lambda-1}{2} = \lambda-2$,we get $\lambda-1 = 2\lambda-4$,which implies $\lambda = 3$. Substituting $\lambda = 3$ into the coordinates of $A$,we get $x = 3$,$y = 3-1 = 2$,and $z = 3$. Therefore,$x+y+z = 3+2+3 = 8$.

$A$ line with direction cosines proportional to $2, 1, 2$ meets the line $L_1$ passing through $(0, -1, 0)$ with direction ratios $1, 1, 1$ at $A(x, y, z)$ and another line $L_2$ at $B(1, 1, 1)$. Then $x+y+z=$

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