A) Find the piercing point of tfracx-11=tfrac y2=tfracz+11…

A) Find the piercing point of $\tfrac{x-1}1=\tfrac y2=\tfrac{z+1}1$ through $x-2y+z+5=0$. B) Write the plane through $M_1(-1,2,-3)$ perpendicular to $\tfrac{x-1}1=\tfrac y2=\tfrac{z+2}1$, and find where the line pierces it.

Explanation of Method

Piercing points of a line and a plane, and the plane through a point perpendicular to a line.

Step 1: Part A

Substituting $x=1+t,\ y=2t,\ z=-1+t$: $5-2t=0$, so $t=\tfrac52$.

\[P_A\left(\tfrac72,\ 5,\ \tfrac32\right)\]

Check: $\tfrac72-10+\tfrac32+5=0$.

Step 2: Part B

The normal is $\{1,2,1\}$: $(x+1)+2(y-2)+(z+3)=0$.

\[x+2y+z=0\]

Substituting $x=1+t,\ y=2t,\ z=-2+t$: $6t-1=0$, so $t=\tfrac16$.

\[P_B\left(\tfrac76,\ \tfrac13,\ -\tfrac{11}6\right)\]

$P_B$ is the foot of the perpendicular from $M_1$ to the line. The distance from $M_1$ to the line is $|M_1P_B|=\sqrt{\tfrac{169+100+49}{36}}=\tfrac{\sqrt{318}}{6}\approx2.97$.

$P_A\left(\tfrac72,5,\tfrac32\right)$

A) The line pierces $x-2y+z+5=0$ at $P_A$ (dashed part on the other side of the plane).

$M_1$ $P_B$

B) The plane $x+2y+z=0$ through $M_1$ is perpendicular to the line; the line pierces it at $P_B\left(\tfrac76,\tfrac13,-\tfrac{11}6\right)$.

Result

\[\boxed{\,P_A\left(\tfrac72,\ 5,\ \tfrac32\right)\,}\]

\[\boxed{\,x+2y+z=0\,}\]

\[\boxed{\,P_B\left(\tfrac76,\ \tfrac13,\ -\tfrac{11}6\right)\,}\]