Where does the line x = 2 − t, y = 3t, z = −1 + 2t intersect the plane 2y + 3z = 6?

Question

form this 2y + 3z = 6 equation i take the x = 0. therefor 2 - t = 0 and t = 2. then i got y = 6 and z = 3 respectively from y = 3t, z = −1 + 2t. but the value does not satisfy the equation 2y + 3z = 6. was my approach wrong ? if then how to solve it ? sorry for my bad English.

Answer 1

According to you, $\;y=3t\;,\;\;z=-1+2t\;,\;\;t\in\Bbb R$ , so that

$$6=2y+3z=2\cdot3t+3(-1+2t)=-3+12t\iff t=\frac34$$

Take it from here.

Answer 2

Suggestion: Express $\,2y + 3z = 6\,$ in terms of $t$, and solve.

$$2(3t) + 3(-1 + 2t) = 6 \iff 6t + 6t-3 = 6 \iff 12t = 9\iff t =\dfrac 34$$

Now find $x, y, z$ at $t = \dfrac 34.$