Solve {l}{text{Thepointontheline}x-2/1=y+3/-2=z+5/-2text{atadistanceof}6text{fromthepoint}(2,-3,-5)text{is:}}{{lllll}{text{(A)}(3,-5,-3)}&{text{(B)}}&{(4,-7,-9)}&{text{(C)}(0,2,-1)}&{text{(D)}quad(-3,5,3)}}

$$\left. \begin{array} { l } { \text { The point on the line } \frac { x - 2 } { 1 } = \frac { y + 3 } { - 2 } = \frac { z + 5 } { - 2 } \text { at a distance of } 6 \text { from the point } ( 2 , - 3 , - 5 ) \text { is : } } \\ { \left. \begin{array} { l l l l l } { \text { (A) } ( 3 , - 5 , - 3 ) } & { \text { (B) } } & { ( 4 , - 7 , - 9 ) } & { \text { (C) } ( 0,2 , - 1 ) } & { \text { (D) } \quad ( - 3,5,3 ) } \end{array} \right. } \end{array} \right.$$

Solve for z

$$z=-1$$

Steps for Solving Linear Equation

Multiply both sides by -2.

$$-2\left(-2\right)=z+5$$

Multiply -2 and -2 to get 4.

$$4=z+5$$

Swap sides so that all variable terms are on the left hand side.

$$z+5=4$$

Subtract 5 from both sides.

$$z=4-5$$

Subtract 5 from 4 to get -1.

$$z=-1$$