Find the line through (8, 1, −5) that intersects and is perpendicular to the line x = −1 + t, y = −3 + t, z = −1 + t. (HINT: If (x0, y0, z0) is the point of intersection, find its coordinates. Enter your answers as a comma-separated list of equations.)

Answer:[tex]x=-6t+8,y=-t+1,z=7t-5[/tex]Step-by-step explanation:A line passing through the point (8, 1, −5) and perpendicular to the line x = −1 + t, y = −3 + t, z = −1 + t.The parallel vector of given line is < 1, 1, 1 >.Let the point of intersection is  (-1+t, -3+t, -1+t).The required line passing through the points (8, 1, −5) and  (-1+t, -3+t, -1+t). So, the vector of required line is< 8-(-1+t), 1-(-3+t), -5-(-1+t) > < 9-t, 4-t, -4-t >The dot product of two perpendicular vectors is 0.Since < 9-t, 4-t, -4-t > and < 1, 1, 1 > are perpendicular vectors, therefore the dot product of these vectors is 0.[tex](9-t)(1)+(4-t)(1)+(-4-t)(1)=0[/tex][tex]9-t+4-t-4-t=0[/tex][tex]9-3t=0[/tex][tex]t=3[/tex]The value of t is 3. So, the point of intersection is [tex](-1+3, -3+3, -1+3)=(2,0,2)[/tex]If a line passes through two points then the equation of line is[tex]\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}[/tex]The line passes through two points (8, 1, −5) and (2,0,2). So, the equation of line is[tex]\frac{x-8}{2-8}=\frac{y-1}{0-1}=\frac{z+5}{2+5}[/tex]Let[tex]\frac{x-8}{-6}=\frac{y-1}{-1}=\frac{z+5}{7}=t[/tex][tex]x=-6t+8,y=-t+1,z=7t-5[/tex]Therefore the parametric equations of the required line are [tex]x=-6t+8,y=-t+1,z=7t-5[/tex].