$(a)\;13\qquad(b)\;12\qquad(c)\;14\qquad(d)\;15$

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Given: $A(2,-1,2)$, $B(5,3,4)$

Direction ratio of the line $AB=(5-2,3+1,4-2)$=(3,4,2)$

Equation of the line $AB$ is given by

$\large\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{2}$$=\lambda$....(i)

Any point on this line is given in terms of $\lambda$ by

$C(3\lambda+2,4\lambda-1,2\lambda+2)$

Let the line $AB$ intersect the plane $x-y+z=5$ at this point $C$

$\Rightarrow\:$ $C$ lies on the plane too.

$\therefore$ $C$ satisfies the equation of the plane.$

$\Rightarrow\:(3\lambda+2)-(4\lambda-1)+(2\lambda+2)=5$

$\Rightarrow\:\lambda=0$

$\therefore\:C(2,-1,2)$

Step 2

We have to find the distance of the point $P(-1,-5,-10)$ from $C(2,-1,2)$

which is given by $\sqrt {2+1)^2+(-1+5)^2+(2+10)^2}=\sqrt {9+16+144}=13$

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