If the line frac{x - 2}{3} = frac{y + 1}{2} = | vedclass

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(A) Let the line be $\frac{x-2}{3}=\frac{y+1}{2}=\frac{z-1}{-1}=\lambda$.Then,any point on the line is given by $x=3\lambda+2, y=2\lambda-1, z=-\lambd

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$2\sqrt{14}$

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(A) Let the line be $\frac{x-2}{3}=\frac{y+1}{2}=\frac{z-1}{-1}=\lambda$.Then,any point on the line is given by $x=3\lambda+2, y=2\lambda-1, z=-\lambda+1$.For the intersection with the plane $2x+3y-z+13=0$:$2(3\lambda+2)+3(2\lambda-1)-(-\lambda+1)+13=0$$6\lambda+4+6\lambda-3+\lambda-1+13=0$$13\lambda+13=0 \implies \lambda=-1$.Substituting $\lambda=-1$,we get $P(-1, -3, 2)$.For the intersection with the plane $3x+y+4z=16$:$3(3\lambda+2)+(2\lambda-1)+4(-\lambda+1)=16$$9\lambda+6+2\lambda-1-4\lambda+4=16$$7\lambda+9=16 \implies 7\lambda=7 \implies \lambda=1$.Substituting $\lambda=1$,we get $Q(5, 1, 0)$.The distance $PQ$ is given by $\sqrt{(5 - (-1))^2 + (1 - (-3))^2 + (0 - 2)^2}$.$PQ = \sqrt{6^2 + 4^2 + (-2)^2} = \sqrt{36 + 16 + 4} = \sqrt{56} = 2\sqrt{14}$.

If the line $\frac{x - 2}{3} = \frac{y + 1}{2} = \frac{z - 1}{-1}$ intersects the plane $2x + 3y - z + 13 = 0$ at a point $P$ and the plane $3x + y + 4z = 16$ at a point $Q$,then $PQ$ is equal to

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