{ "problem": "Find the point in the plane $3x - 4y + 5z = 30$ that is closest to the point $(1,2,3).$", "level": "Level 4", "type": "Precalculus", "solution": "Let $A = (1,2,3),$ and let $P$ be the point in the plane which is closest to $A.$\n\n[asy]\nimport three;\n\nsize(180);\ncurrentprojection = perspective(6,3,2);\n\ntriple I = (1,0,0), J = (0,1,0), K = (0,0,1), O = (0,0,0);\ntriple A = (0,1.8,1), P = (0,1.8,0);\n\ndraw(surface((2*I + 3*J)--(2*I - 1*J)--(-2*I - 1*J)--(-2*I + 3*J)--cycle),paleyellow,nolight);\ndraw((2*I + 3*J)--(2*I - 1*J)--(-2*I - 1*J)--(-2*I + 3*J)--cycle);\ndraw(A--P);\n\ndot(\"$A$\", A, N);\ndot(\"$P$\", P, E);\n[/asy]\n\nThen $\\overrightarrow{AP}$ is a multiple of the normal vector of the plane, which is $\\begin{pmatrix} 3 \\\\ -4 \\\\ 5 \\end{pmatrix}.$ Thus,\n\\[\\overrightarrow{AP} = t \\begin{pmatrix} 3 \\\\ -4 \\\\ 5 \\end{pmatrix}\\]for some scalar $t.$ This means point $P$ is of the form $(1 + 3t, 2 - 4t, 3 + 5t).$ But we also know $P$ lies in the plane $3x - 4y + 5z = 30,$ so\n\\[3(1 + 3t) - 4(2 - 4t) + 5(3 + 5t) = 30.\\]Solving for $t,$ we find $t = \\frac{2}{5}.$ Therefore, $P = \\boxed{\\left( \\frac{11}{5}, \\frac{2}{5}, 5 \\right)}.$" }